balance chemical equation question answer

Balancing Chemical Equations: Explanation, Review, and Examples

• The Albert Team
• Last Updated On: March 14, 2023

Of all the skills to know about in chemistry, balancing chemical equations is perhaps the most important to master. So many parts of chemistry depend on this vital skill, including stoichiometry, reaction analysis, and lab work. This comprehensive guide will show you the steps to balance even the most challenging reactions and will walk you through a series of examples, from simple to complex.

The Key to Balancing Chemical Equations

The ultimate goal for balancing chemical equations is to make both sides of the reaction, the reactants and the products, equal in the number of atoms per element. This stems from the universal law of the conservation of mass, which states that matter can neither be created nor destroyed. So, if we start with ten atoms of oxygen before a reaction, we need to end up with ten atoms of oxygen after a reaction. This means that chemical reactions do not change the actual building blocks of matter; rather, they just change the arrangement of the blocks. An easy way to understand this is to picture a house made of blocks. We can break the house apart and build an airplane, but the color and shape of the actual blocks do not change.

But how do we go about balancing these equations? We know that the number of atoms of each element needs to be the same on both sides of the equation, so it is just a matter of finding the correct coefficients (numbers in front of each molecule) to make that happen. It is best to start with the atom that shows up the least number of times on one side, and balance that first. Then, move on to the atom that shows up the second least number of times, and so on. In the end, make sure to count the number of atoms of each element on each side again, just to be sure.

Example of Balancing a Chemical Equation

Let’s illustrate this with an example by balancing this chemical equation:

P 4 O 10 + H 2 O → H 3 PO 4

First, let’s look at the element that appears least often. Notice that oxygen occurs twice on the left-hand side, so that is not a good element to start out with. We could either start with phosphorus or hydrogen, so let’s start with phosphorus. There are four atoms of phosphorus on the left-hand side, but only one on the right-hand side. So, we can put the coefficient of 4 on the molecule that has phosphorous on the right-hand side to balance them out.

P 4 O 10 + H 2 O → 4 H 3 PO 4

Now we can check hydrogen. We still want to avoid balancing oxygen, because it occurs in more than one molecule on the left-hand side. It is easiest to start with molecules that only appear once on each side. So, there are two molecules of hydrogen on the left-hand side and twelve on the right-hand side (notice that there are three per molecule of H 3 PO 4 , and we have four molecules). So, to balance those out, we have to put a six in front of H 2 O on the left.

P 4 O 10 + 6 H 2 O → 4 H 3 PO 4

At this point, we can check the oxygens to see if they balance. On the left, we have ten atoms of oxygen from P 4 O 10 and six from H 2 O for a total of 16. On the right, we have 16 as well (four per molecule, with four molecules). So, oxygen is already balanced. This gives us the final balanced equation of

Balancing Chemical Equations Practice Problems

Try to balance these ten equations on your own, then check the answers below. They range in difficulty level, so don’t get discouraged if some of them seem too hard. Just remember to start with the element that shows up the least, and proceed from there. The best way to approach these problems is slowly and systematically. Looking at everything at once can easily get overwhelming. Good luck!

• CO 2 + H 2 O → C 6 H 12 O 6 + O 2
• SiCl 4 + H 2 O → H 4 SiO 4 + HCl
• Al + HCl → AlCl 3 + H 2
• Na 2 CO 3 + HCl → NaCl + H 2 O + CO 2
• C 7 H 6 O 2 + O 2 → CO 2 + H 2 O
• Fe 2 (SO 4 ) 3 + KOH → K 2 SO 4 + Fe(OH) 3
• Ca 3 (PO 4 ) 2 + SiO 2 → P 4 O 10 + CaSiO 3
• KClO 3 → KClO 4 + KCl
• Al 2 (SO 4 ) 3 + Ca(OH) 2 → Al(OH) 3 + CaSO 4
• H 2 SO 4 + HI → H 2 S + I 2 + H 2 O

Complete Solutions:

1. co 2 + h 2 o → c 6 h 12 o 6 + o 2.

The first step to balancing chemical equations is to focus on elements that only appear once on each side of the equation. Here, both carbon and hydrogen fit this requirement. So, we will start with carbon. There is only one atom of carbon on the left-hand side, but six on the right-hand side. So, we add a coefficient of six to the carbon-containing molecule on the left.

6CO 2 + H 2 O → C 6 H 12 O 6 + O 2

Next, let’s look at hydrogen. There are two hydrogen atoms on the left and twelve on the right. So, we will add a coefficient of six on the hydrogen-containing molecule on the left.

6CO 2 + 6H 2 O → C 6 H 12 O 6 + O 2

Now, it is time to check the oxygen. There are a total of 18 oxygen molecules on the left (6×2 + 6×1). On the right, there are eight oxygen molecules. Now, we have two options to even out the right-hand side: We can either multiply C 6 H 12 O 6 or O 2 by a coefficient. However, if we change C 6 H 12 O 6 , the coefficients for everything else on the left-hand side will also have to change, because we will be changing the number of carbon and hydrogen atoms. To prevent this, it usually helps to only change the molecule containing the fewest elements; in this case, the O 2 . So, we can add a coefficient of six to the O 2 on the right. Our final answer will be:

6CO 2 + 6H 2 O → C 6 H 12 O 6 + 6O 2

2. SiCl 4 + H 2 O → H 4 SiO 4 + HCl

The only element that occurs more than once on the same side of the equation here is hydrogen, so we can start with any other element. Let’s start by looking at silicon. Notice that there is only one atom of silicon on either side, so we do not need to add any coefficients yet. Next, let’s look at chlorine. There are four chlorine atoms on the left side and only one on the right. So, we will add a coefficient of four on the right.

SiCl 4 + H 2 O → H 4 SiO 4 + 4HCl

Next, let’s look at oxygen. Remember that we first want to analyze all the elements that only occur once on one side of the equation. There is only one oxygen atom on the left, but four on the right. So, we will add a coefficient of four on the left-hand side of the equation.

SiCl 4 + 4H 2 O → H 4 SiO 4 + 4HCl

We are almost done! Now, we just have to check the number of hydrogen atoms on each side. The left has eight and the right also has eight, so we are done. Our final answer is

As always, make sure to double-check that the number of atoms of each element balances on each side before continuing.

3. Al + HCl → AlCl 3 + H 2

This problem is a bit tricky, so be careful. Whenever a single atom is alone on either side of the equation, it is easiest to start with that element. So, we will start by counting the aluminum atoms on both sides. There is one on the left and one on the right, so we do not need to add any coefficients yet. Next, let’s look at hydrogen. There is also one on the left, but two on the right. So, we will add a coefficient of two on the left.

Al + 2HCl → AlCl 3 + H 2

Next, we will look at chlorine. There are now two on the left, but three on the right. Now, this is not as straightforward as just adding a coefficient to one side. We need the number of chlorine atoms to be equal on both sides, so we need to get two and three to be equal. We can accomplish this by finding the lowest common multiple. In this case, we can multiply two by three and three by two to get the lowest common multiple of six. So, we will multiply 2HCl by three and AlCl 3 by two:

Al + 6HCl → 2AlCl 3 + H 2

We have looked at all the elements, so it is easy to say that we are done. However, always make sure to double-check. In this case, because we added a coefficient to the aluminum-containing molecule on the right-hand side, aluminum is no longer balanced. There is one on the left but two on the right. So, we will add one more coefficient.

2Al + 6HCl → 2AlCl 3 + H 2

We are not quite done yet. Looking over the equation one final time, we see that hydrogen has also been unbalanced. There are six on the left but two on the right. So, with one final adjustment, we get our final answer:

2Al + 6HCl → 2AlCl 3 + 3H 2

4. Na 2 CO 3 + HCl → NaCl + H 2 O + CO 2

Hopefully, by this point, balancing equations is becoming easier and you are getting the hang of it. Looking at sodium, we see that it occurs twice on the left, but once on the right. So, we can add our first coefficient to the NaCl on the right.

Na 2 CO 3 + HCl → 2NaCl + H 2 O + CO 2

Next, let’s look at carbon. There is one on the left and one on the right, so there are no coefficients to add. Since oxygen occurs in more than one place on the left, we will save it for last. Instead, look at hydrogen. There is one on the left and two on the right, so we will add a coefficient to the left.

Na 2 CO 3 + 2HCl → 2NaCl + H 2 O + CO 2

Then, looking at chlorine, we see that it is already balanced with two on each side. Now we can go back to look at oxygen. There are three on the left and three on the right, so our final answer is

5. C 7 H 6 O 2 + O 2 → CO 2 + H 2 O

We can start balancing this equation by looking at either carbon or hydrogen. Looking at carbon, we see that there are seven atoms on the left and only one on the right. So, we can add a coefficient of seven on the right.

C 7 H 6 O 2 + O 2 → 7CO 2 + H 2 O

Then, for hydrogen, there are six atoms on the left and two on the right. So, we will add a coefficient of three on the right.

C 7 H 6 O 2 + O 2 → 7CO 2 + 3H 2 O

Now, for oxygen, things will get a little tricky. Oxygen occurs in every molecule in the equation, so we have to be very careful when balancing it. There are four atoms of oxygen on the left and 17 on the right. There is no obvious way to balance these numbers, so we must use a little trick: fractions. Now, when balancing chemical equations, we cannot include fractions as it is not proper form, but it sometimes helps to use them to solve the problem. Also, try to avoid over-manipulating organic molecules. You can easily identify organic molecules, otherwise known as CHO molecules, because they are made up of only carbon, hydrogen, and oxygen. We don’t like to work with these molecules, because they are rather complex. Also, larger molecules tend to be more stable than smaller molecules, and less likely to react in large quantities.

So, to balance out the four and seventeen, we can multiply the O 2 on the left by 7.5. That will give us

C 7 H 6 O 2 + 7.5O 2 → 7CO 2 + 3H 2 O

Remember, fractions (and decimals) are not allowed in formal balanced equations, so multiply everything by two to get integer values. Our final answer is now

2C 7 H 6 O 2 + 15O 2 → 14CO 2 + 6H 2 O

6. Fe 2 (SO 4 ) 3 + KOH → K 2 SO 4 + Fe(OH) 3­-

We can start by balancing the iron on both sides. The left has two while the right only has one. So, we will add a coefficient of two to the right.

Fe 2 (SO 4 ) 3 + KOH → K 2 SO 4 + 2Fe(OH) 3­-

Then, we can look at sulfur. There are three on the left, but only one on the right. So, we will add a coefficient of three to the right-hand side.

Fe 2 (SO 4 ) 3 + KOH → 3K 2 SO 4 + 2Fe(OH) 3­-

We are almost done. All that is left is to balance the potassium. There is one atom on the left and six on the right, so we can balance these by adding a coefficient of six. Our final answer, then, is

Fe 2 (SO 4 ) 3 + 6KOH → 3K 2 SO 4 + 2Fe(OH) 3­-

7. Ca 3 (PO 4 ) 2 + SiO 2 → P 4 O 10 + CaSiO 3

Looking at calcium, we see that there are three on the left and one on the right, so we can add a coefficient of three on the right to balance them out.

Ca 3 (PO 4 ) 2 + SiO 2 → P 4 O 10 + 3CaSiO 3

Then, for phosphorus, we see that there are two on the left and four on the right. To balance these, add a coefficient of two on the left.

2Ca 3 (PO 4 ) 2 + SiO 2 → P 4 O 10 + 3CaSiO 3

Notice that by doing so, we changed the number of calcium atoms on the left. Every time you add a coefficient, double check to see if the step affects any elements you have already balanced. In this case, the number of calcium atoms on the left has increased to six while it is still three on the right, so we can change the coefficient on the right to reflect this change.

2Ca 3 (PO 4 ) 2 + SiO 2 → P 4 O 10 + 6CaSiO 3

Since oxygen occurs in every molecule in the equation, we will skip it for now. Focusing on silicon, we see that there is one on the left, but six on the right, so we can add a coefficient to the left.

2Ca 3 (PO 4 ) 2 + 6SiO 2 → P 4 O 10 + 6CaSiO 3

Now, we will check the number of oxygen atoms on each side. The left has 28 atoms and the right also has 28. So, after checking that all the other atoms are the same on both sides as well, we get a final answer of

8. KClO 3 → KClO 4 + KCl

This problem is particularly tricky because every atom, except oxygen, occurs in every molecule in the equation. So, since oxygen appears the least number of times, we will start there. There are three on the left and four on the right. To balance these, we find the lowest common multiple; in this case, 12. By adding a coefficient of four on the left and three on the right, we can balance the oxygens.

4KClO 3 → 3KClO 4 + KCl

Now, we can check potassium and chlorine. There are four potassium molecules on the left and four on the right, so they are balanced. Chlorine is also balanced, with four on each side, so we are finished, with a final answer of

9. Al 2 (SO 4 ) 3 + Ca(OH) 2 → Al(OH) 3 + CaSO 4

We can start here by balancing the aluminum atoms on both sides. The left has two molecules while the right only has one, so we will add a coefficient of two on the right.

Al 2 (SO 4 ) 3 + Ca(OH) 2 → 2Al(OH) 3 + CaSO 4

Now, we can check sulfur. There are three on the left and only one on the right, so adding a coefficient of three will balance these.

Al 2 (SO 4 ) 3 + Ca(OH) 2 → 2Al(OH) 3 + 3CaSO 4

Moving right along to calcium, there is only one on the left but three on the right, so we should add a coefficient of three.

Al 2 (SO 4 ) 3 + 3Ca(OH) 2 → 2Al(OH) 3 + 3CaSO 4

Double-checking all the atoms, we see that all the elements are balanced, so our final equation is

10. H 2 SO 4 + HI → H 2 S + I 2 + H 2 O

Since hydrogen occurs more than once on the left, we will temporarily skip it and move to sulfur. There is one atom on the left and one on the right, so there is nothing to balance yet. Looking at oxygen, there are four on the left and one on the right, so we can add a coefficient of four to balance them.

H 2 SO 4 + HI → H 2 S + I 2 + 4H 2 O

There is only one iodine on the left and two on the right, so a simple coefficient change can balance those.

H 2 SO 4 + 2HI → H 2 S + I 2 + 4H 2 O

Now, we can look at the most challenging element: hydrogen. On the left, there are four and on the right, there are ten. So, we know we have to change the coefficient of either H 2 SO 4 or HI. We want to change something that will require the least amount of tweaking afterwards, so we will change the coefficient of HI. To get the left-hand side to have ten atoms of hydrogen, we need HI to have eight atoms of hydrogen, since H 2 SO 4 already has two. So, we will change the coefficient from 2 to 8.

H 2 SO 4 + 8HI → H 2 S + I 2 + 4H 2 O

However, this also changes the balance for iodine. There are now eight on the left, but only two on the right. To fix this, we will add a coefficient of 4 on the right. After checking that everything else balances out as well, we get a final answer of

H 2 SO 4 + 8HI → H 2 S + 4I 2 + 4H 2 O

As with most skills, practice makes perfect when balancing chemical equations. Keep working hard and try to do as many problems as you can to help you hone your balancing skills.

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Balancing Chemical Equations Quiz – Questions and Answers

Here is a ten-question balancing chemical equations quiz. Each question presents an unbalanced equation. Select the balanced equation. Find the answer key below the questions. Remember, the number of each type of atoms is the same on both sides of the reaction arrow when an equation is balanced.

Balancing Chemical Equations Quiz Questions

(1) __ Na₃PO₄ + __ HCl → __ NaCl + __ H₃PO₄

• Na₃PO₄ + HCl → NaCl + H₃PO₄
• Na₃PO₄ +3 HCl → 3 NaCl + H₃PO₄
• 3 Na₃PO₄ + HCl → 3 NaCl + H₃PO₄
• Na₃PO₄ + 3 HCl → NaCl + H₃PO₄

(2) __ TiCl₄ + __ H₂O → __ TiO₂ + __ HCl

• TiCl₄ + 2 H₂O → TiO₂ + 2 HCl
• TiCl₄ + 2 H₂O → TiO₂ + 4 HCl
• 2 TiCl₄ + H₂O → 2 TiO₂ + HCl
• TiCl₄ + 4 H₂O → TiO₂ + 4 HCl

(3) __ C₂H₆O + __ O₂ → __ CO₂ + __ H₂O

• 2 C₂H₆O + O₂ → CO₂ + 3 H₂O
• C₂H₆O + 3 O₂ → 2 CO₂ + 3 H₂O
• C₂H₆O + 2 O₂ → 3 CO₂ + 3 H₂O
• 2 C₂H₆O + O₂ → 2 CO₂ + H₂O

(4) __ FeS + __ O₂ → __ Fe₂O₃ + __ SO₂

• 2 FeS + O₂ → 2 Fe₂O₃ + 2 SO₂
• FeS + O₂ → Fe₂O₃ + SO₂
• 3 FeS + 2 O₂ → 3 Fe₂O₃ + 2 SO₂
• 4 FeS + 7 O₂ → 2 Fe₂O₃ + 4 SO₂

(5) __ CaCl₂ + __ Na₃PO₄ → __ Ca₃(PO₄)₂ + __ NaCl

• 2 CaCl₂ + 2 Na₃PO₄ → 2 Ca₃(PO₄)₂ + NaCl
• CaCl₂ + Na₃PO₄ → Ca₃(PO₄)₂ + NaCl
• 3 CaCl₂ + 2 Na₃PO₄ → Ca₃(PO₄)₂ + 6 NaCl
• 3 CaCl₂ + 2 Na₃PO₄ → Ca₃(PO₄)₂ + 3 NaCl

(6) __ KOH + __ H₃PO₄ → __ K₃PO₄ + __ H₂O

• 3 KOH + 2 H₃PO₄ → 3 K₃PO₄ + 3 H₂O
• KOH + H₃PO₄ → K₃PO₄ + 6 H₂O
• 3 KOH + H₃PO₄ → K₃PO₄ + 3 H₂O
• 2 KOH + H₃PO₄ → K₃PO₄ + H₂O

(7) __ AgI + __ Na₂S → __ Ag₂S + __ NaI

• AgI + Na₂S → Ag₂S + 2 NaI
• 2 AgI + 2 Na₂S → 2 Ag₂S + 2 NaI
• 2 AgI + 2 Na₂S → Ag₂S + 2 NaI
• 2 AgI + Na₂S → Ag₂S + 2 NaI

(8) __ Ba₃N₂ + __ H₂O → __ Ba(OH)₂ + __ NH₃

• 2 Ba₃N₂ + 3 H₂O → 2 Ba(OH)₂ + 2 NH₃
• Ba₃N₂ + 6 H₂O → 3 Ba(OH)₂ + 2 NH₃
• Ba₃N₂ + 3 H₂O → 3 Ba(OH)₂ + 2 NH₃
• Ba₃N₂ + H₂O → Ba(OH)₂ + NH₃

(9) ___ SnO₂ + ___ H₂ → ___ Sn + ___ H₂O

• 2 SnO₂ + 2 H₂ → 2 Sn + H₂O
• SnO₂ + 2 H₂ → Sn + 2 H₂O
• SnO₂ + H₂ → Sn + 2 H₂O
• 2 SnO₂ + H₂ → 2Sn + H₂O

(10) __ KNO₃ + __ H₂CO₃ → __ K₂CO₃ + __ HNO₃

• 2 KNO₃ + H₂CO₃ → K₂CO₃ + 2 HNO₃
• 2 KNO₃ + 2 H₂CO₃ → K₂CO₃ + 2 HNO₃
• KNO₃ + H₂CO₃ → K₂CO₃ + HNO₃
• 2 KNO₃ + 2 H₂CO₃ → K₂CO₃ + 3 HNO₃

Balancing Chemical Equations Quiz Answers

Learn more about balancing equations.

If you had trouble with this quiz, it’s a good idea to review the steps of balancing chemical equations.

• Guide to Balancing Chemical Equations – Learn the steps for balancing chemical equations and see how to check your work.
• Examples of Balanced Equations – Not sure what a balanced equation looks like? Here are some examples.
• Practice Worksheet – If you want more practice, try worksheets! Print worksheets out and improve your skills.

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How to Balance Chemical Equations

Last Updated: October 13, 2023 Fact Checked

This article was co-authored by Bess Ruff, MA . Bess Ruff is a Geography PhD student at Florida State University. She received her MA in Environmental Science and Management from the University of California, Santa Barbara in 2016. She has conducted survey work for marine spatial planning projects in the Caribbean and provided research support as a graduate fellow for the Sustainable Fisheries Group. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 4,829,946 times.

Taking a dive into the world of chemical equations? These problems can seem tricky at a glance, but they’re easy to figure out once you learn the basic steps and rules to balancing them. Not to worry; we’ll walk you through exactly how to figure out just about any problem, no matter how many atoms and molecules you're working with. Dealing with especially complex equations? We’ve got you covered there, too—scroll to section 2 for a handy tutorial on solving trickier equations with an algebraic balance.

Doing a Traditional Balance

• C 3 H 8 + O 2 --> H 2 O + CO 2
• This reaction occurs when propane (C 3 H 8 ) is burned in the presence of oxygen to produce water and carbon dioxide.

• For example, you have 3 oxygen atoms on the right side, but that total results from addition.
• Left side: 3 carbon (C3), 8 hydrogen (H8) and 2 oxygen (O2).
• Right side: 1 carbon (C), 2 hydrogen (H2) and 3 oxygen (O + O2).

• You'll need to recount your atoms before balancing the hydrogen and oxygen, as you'll likely need to use coefficients to balance the other atoms in the equation.

• C 3 H 8 + O 2 --> H 2 O + 3 CO 2
• The coefficient 3 in front of carbon on the right side indicates 3 carbon atoms just as the subscript 3 on the left side indicates 3 carbon atoms.
• In a chemical equation, you can change coefficients, but you must never alter the subscripts.

• C 3 H 8 + O 2 --> 4 H 2 O + 3CO 2
• On the right side, you now added a 4 as the coefficient because the subscript showed that you already had 2 hydrogen atoms.
• When you multiply the coefficient 4 times by the subscript 2, you end up with 8.

• Add a coefficient of 5 to the oxygen molecule on the left side of the equation. You now have 10 oxygen atoms on each side.

• The carbon, hydrogen, and oxygen atoms are balanced. Your equation is complete.
• The other 6 atoms of oxygen come from 3CO 2 .(3x2=6 atoms of oxygen+ the other 4=10)

Completing an Algebraic Balance

This method, also known as Bottomley's method, is especially useful for more complex reactions, although it does take a bit longer.

• PCl 5 + H 2 O --> H 3 PO 4 + HCl

• a PCl 5 + b H 2 O --> c H 3 PO 4 + d HCl

• On the left side, there are 2 b atoms of hydrogen (2 for every molecule of H 2 O), while on the right side, there are 3 c + d atoms of hydrogen (3 for every molecule of H 3 PO 4 and 1 for every molecule of HCl). Since the number of atoms of hydrogen has to be equal on both sides, 2 b must be equal to 3 c + d .
• Cl: 5 a = d
• H: 2 b =3 c + d

• To quickly do this, take one variable and assign a value to it. Let's make a = 1. Then start solving the system of equations to get the following values:
• Since P: a = c, we know that c = 1.
• Since Cl: 5a = d, we know that d = 5
• 2b = 3(1) + 5
• If the value you assigned returns fractional values, just multiply all values by the least common multiple (LCM) of the denominators to get rid of the fractions. If there is only one fraction, multiply all values by that values denominator.
• If the value you assigned returns coefficients that have a greatest common factor (GCF), simplify the chemical equation by dividing each value by the GCF.

Community Q&A

Video . By using this service, some information may be shared with YouTube.

• Remember to simplify! If all of your coefficients can be divided by the same number, do so to get the simplest result. Thanks Helpful 36 Not Helpful 10
• If you're stuck, you can type the equation into the online balancer to balance it. Just remember that you won't have access to an online balancer when you're taking an exam, so don't become dependent on it. Thanks Helpful 47 Not Helpful 23
• To get rid of fractions, multiply the entire equation (both the left and right sides) by the number in the denominator of your fraction. Thanks Helpful 23 Not Helpful 10

• During the balancing process, you may use fractions to assist you, but the equation is not balanced as long as there are still coefficients using fractions. You never make half of a molecule or half of an atom in a chemical reaction. Thanks Helpful 19 Not Helpful 11

You Might Also Like

• ↑ https://www.khanacademy.org/science/ap-biology/chemistry-of-life/elements-of-life/a/matter-elements-atoms-article
• ↑ https://www.mathsisfun.com/algebra/add-subtract-balance.html
• ↑ https://www.khanacademy.org/science/chemistry/chemical-reactions-stoichiome/balancing-chemical-equations/v/balancing-chemical-equations-introduction
• ↑ https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Redox_Chemistry/Balancing_Redox_reactions
• ↑ https://www.mathsisfun.com/algebra/completing-square.html

About This Article

To balance a chemical equation, first write out your given formula with the reactants on the left of the arrow and the products on the right. For example, your equation should look something like "H2 + O2 → H2O." Count the number of atoms in each element on each side of the equation and list them under that side. For the equation H2 + O2 → H2O, there are 2 hydrogen atoms being added to 2 oxygen atoms on the left, so you would write "H=2" and "O=2" under the left side. There are 2 hydrogen atoms and 1 oxygen atom on the right, so you would write "H=2" and "O=1" under the right side. Since the number of atoms in each element isn't identical on both sides, the equation is not balanced. To balance the equation, you'll need to add coefficients to change the number of atoms on one side to match the other. For the equation H2 + O2 → H2O, you would add the coefficient 2 before H2O on the right side so that there are 2 oxygen atoms on each side of the equation, like H2 + O2 → 2H2O. However, subscripts can't be changed and are always multiplied by the coefficient, which means there are now 4 hydrogen atoms on the right side of the equation and only 2 hydrogen atoms on the left side. To balance this, add the coefficient 2 before H2 on the left side of the equation so there are 4 hydrogen atoms on each side, like 2H2 + O2 → 2H2O. Now the number of atoms in each element is the same on both sides of the equation, so the equation is balanced. Remember that if there's no coefficient in front of an element, it's assumed that the coefficient is 1. To learn how to balance chemical equations algebraically, scroll down! Did this summary help you? Yes No

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Balancing Equations Practice Quiz

Here are 10 unbalanced equations. Select the correct balanced equation.

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• 2 SnO₂ + 2 H₂ → 2 Sn + H₂O
• SnO₂ + 2 H₂ → Sn + 2 H₂O
• SnO₂ + H₂ → Sn + 2 H₂O
• 2 SnO₂ + H₂ → 2Sn + H₂O
• 3 KOH + 2 H₃PO₄ → 3 K₃PO₄ + 3 H₂O
• KOH + H₃PO₄ → K₃PO₄ + 6 H₂O
• 3 KOH + H₃PO₄ → K₃PO₄ + 3 H₂O
• 2 KOH + H₃PO₄ → K₃PO₄ + H₂O
• 2 KNO₃ + H₂CO₃ → K₂CO₃ + 2 HNO₃
• 2 KNO₃ + 2 H₂CO₃ → K₂CO₃ + 2 HNO₃
• KNO₃ + H₂CO₃ → K₂CO₃ + HNO₃
• 2 KNO₃ + 2 H₂CO₃ → K₂CO₃ + 3 HNO₃
• AgI + Na₂S → Ag₂S + 2 NaI
• 2 AgI + 2 Na₂S → 2 Ag₂S + 2 NaI
• 2 AgI + 2 Na₂S → Ag₂S + 2 NaI
• 2 AgI + Na₂S → Ag₂S + 2 NaI
• 2 Ba₃N₂ + 3 H₂O → 2 Ba(OH)₂ + 2 NH₃
• Ba₃N₂ + 6 H₂O → 3 Ba(OH)₂ + 2 NH₃
• Ba₃N₂ + 3 H₂O → 3 Ba(OH)₂ + 2 NH₃
• Ba₃N₂ + H₂O → Ba(OH)₂ + NH₃
• 2 CaCl₂ + 2 Na₃PO₄ → 2 Ca₃(PO₄)₂ + NaCl
• CaCl₂ + Na₃PO₄ → Ca₃(PO₄)₂ + NaCl
• 3 CaCl₂ + 2 Na₃PO₄ → Ca₃(PO₄)₂ + 6 NaCl
• 3 CaCl₂ + 2 Na₃PO₄ → Ca₃(PO₄)₂ + 3 NaCl
• 2 FeS + O₂ → 2 Fe₂O₃ + 2 SO₂
• FeS + O₂ → Fe₂O₃ + SO₂
• 3 FeS + 2 O₂ → 3 Fe₂O₃ + 2 SO₂
• 4 FeS + 7 O₂ → 2 Fe₂O₃ + 4 SO₂
• 2 C₂H₆O + O₂ → CO₂ + 3 H₂O
• C₂H₆O + 3 O₂ → 2 CO₂ + 3 H₂O
• C₂H₆O + 2 O₂ → 3 CO₂ + 3 H₂O
• 2 C₂H₆O + O₂ → 2 CO₂ + H₂O
• TiCl₄ + 2 H₂O → TiO₂ + 2 HCl
• TiCl₄ + 2 H₂O → TiO₂ + 4 HCl
• 2 TiCl₄ + H₂O → 2 TiO₂ + HCl
• TiCl₄ + 4 H₂O → TiO₂ + 4 HCl
• Na₃PO₄ + HCl → NaCl + H₃PO₄
• Na₃PO₄ +3 HCl → 3 NaCl + H₃PO₄
• 3 Na₃PO₄ + HCl → 3 NaCl + H₃PO₄
• Na₃PO₄ + 3 HCl → NaCl + H₃PO₄

Good job! You completed the quiz, so you got practice balancing equations. However, you missed some questions, so you might want to review the steps to balancing equations or print free practice worksheets . If you feel ready to move on, learn about mass relations in balanced equations .

Are you ready to try another quiz? See how well you can do metric unit conversions .

Great job! You did so well on this quiz you could tutor others on how to balance equations! If you're feeling a bit shaky on all the steps and details, you can review the simple method of balancing equations . Otherwise, you may wish to review how to balance oxidation-reduction or redox reactions  or move on to understanding mole relations in balanced equations .

Ready to try another quiz? See whether you're safe in the chemistry lab (or are an accident waiting to happen).

• Balancing Equations Test Questions
• How to Balance Chemical Equations
• Balancing Chemical Equations
• Examples of 10 Balanced Chemical Equations
• Balanced Equation Definition and Examples
• Mole Relations in Balanced Equations
• 5 Steps for Balancing Chemical Equations
• Equation for the Decomposition of Sodium Bicarbonate (Baking Soda)
• Example Problem of Mass Relations in Balanced Equations
• How to Balance Equations - Printable Worksheets
• How to Balance Net Ionic Equations
• Balance Redox Reaction Example Problem
• Balance Redox Reaction in Basic Solution Example Problem
• How to Balance Redox Reactions
• Equilibrium Constants Practice Test

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Balancing Chemical Equations with Ease (100 Free Worksheets)

• October 2, 2019

In this article, you will learn about how to balance chemical equations easily with simple steps. Feel free to download our free worksheets with answers for your practice.

Parts of a Balanced Chemical Equation

Before you start balancing chemical equations, it is important that you become firmly acquainted with the various part of one. Every balanced chemical equation consists of two parts: the reactant side and the product side. Both of these sides are separated by the means of an arrow.

On the left side of the arrow, you will find the reactant side. This side represents the elements which are used for initiating the experiment. On the right side of the arrow, we have the product side. This side is used to display the elements or compounds which are generated from the chemical process.

The Need for Balancing Chemical Equations

Ever since you started learning about the field of Chemistry, your teachers might often have stressed upon the importance of balancing chemical equations. But have you ever pondered on the reason behind this? Why do you need to respect the law of the conservation of mass? Quite simply, you need to balance your equations to follow the law of conservation of mass.

Law of Conservation of Mass: According to the law of conservation of mass, the mass of products that are derived from a chemical equation should mandatorily equal the mass of the reactants.

Minding the law of conservation of mass while balancing equations is quite important. Not only does it help you to prevent errors, but it also assists scientists in knowing the quantities of reactants to create a particular product that they want to make. Moreover, the law also helps chemical manufacturers to increase the efficiencies of their processes.

When Should You Start To Balance A Chemical Equation?

As soon as you derive a chemical equation stating the reactants and the products, check out if the number of atoms on either side are equal to each other or not. In the event that you detect the numbers to be uncommon, rest assured, you should start balancing the elements and compounds on either side of the chemical equation.

How Should You Balance A Chemical Equation?

The primary aspect that you need to keep in mind while balancing a chemical equation is this; the entire process is completely based on trial and error. When you start to balance a particular chemical equation, you will need to go through several processes before you stumble upon the right coefficients to balance the number of atoms.

Another aspect that you need to remember is that balancing chemical equations requires a lot of practice. Once you perfect the practice of balancing, you can become completely reliant on your intuition to lead you through the complete process.

While balancing your equations, you need to follow certain simple stems. Here’s what you need to do:

• Start by counting the number of atoms, present for each element on the side of the reactants as well as the products.
• When you find out that certain elements are not balanced, place the required coefficient that is needed to balance the elements.
• Once you are done with this, check out if the number of atoms for the other elements is also equal on both the sides.
• Repeat the process until you find out that all the elements on both the sides of the chemical equations are balanced.

As aforementioned, the process in itself is quite simple. However, it takes significant practice before can start to balance these equations completely with your intuition.

An Easy Example To Get You Started With

Now that you know the steps, you are wholly capable of balancing chemical equations. Let’s solve some of them, shall we? With the help of above-mentioned steps and a practical example, you will be better able to understand how the entire process works.

Do not feel anxious if you feel that you are still not ready to solve these problems. With our method, even your toddler sibling will be able to understand how chemical equations are balanced. And if still feel a tad bit confused after solving all these equations, try to solve a few more of such problems. Remember what we advised in a previous section: You will need significant practice before you can confidently start to balance these equations with your intuition.

Let’s start with this example. This equation represents a reaction between two Iron Oxide (Fe 2 O 3 ) and Carbon (C). The products formed are Iron (Fe) and Carbon Dioxide (CO 2 ).

Fe 2 O 3 + C        →       Fe + CO 2

Alright, so we have our equation. Let’s begin to balance the equation with the help of the steps mentioned above.

Step 1: Start by counting the number of atoms present, for each element on the side of the reactants as well as the products.

On the reactants side, we have:

• 2 atoms of Fe
• 3 atoms of O
• 1 atom of C

On the product side, we have:

• 1 atom of Fe
• 2 atoms of O

By comparing the number of atoms present for each element on each side, you might have determined that the reaction is obviously not balanced. Therefore, let’s move on to Step 2.

Step 2: When you find out that certain elements are not balanced, place the required coefficient that is needed to balance the elements.

Let’s start by balancing the oxygen atoms. To do this, make the oxygen atoms as six on either side of the chemical equation.

2Fe 2 O 3 + C      →         Fe + 3CO 2

On towards the next step now.

Step 3: Once you are done with this, check out if the number of atoms for the other elements is also equal on both the sides.

Now that we have an equal number of oxygen atoms on either side of the equation, let’s check out if the other elements of the equation are equal or not.

2Fe 2 O 3 + C      →        Fe + 3CO 2

On the reactant side, we have:

• 4 atoms of Fe.
• 6 atoms of O.
• 1 atom of C.
• 1 atom of Fe.
• 3 atoms of C.

As you can see, the elements of iron and carbon are still not balanced. Therefore, it is time that we move on to the 4 th step.

Step 4: Repeat the process until you find out that all the elements on both the sides of the chemical equations are balanced.

Alright, let’s start balancing the equation again and this time, let’s balance the number of iron atoms first. On the reactant side, we have 4 atoms of Fe while the product side has 1 atom of Fe. To balance them, we need to place 4 atoms of Fe on the product side.

2Fe 2 O 3 + C      →        4Fe + 3CO 2

Now, on the reactant side we have:

And, on the product side, we have:

The only element that remains to be balanced now is carbon. This can be easily done considering the fact that carbon exists only in a singular form on the reactant side. In order to correct this, we need to place 3 atoms of carbon on the reactant side. The chemical reaction, hence, will turn out to be:

2Fe 2 O 3 + 3C    →         4Fe + 3CO 2

And there you go. Perfectly balanced as all things should be (Yes. We are Marvel fanboys as well).

Essential Tips for Beginners

As you become further acquainted with balancing chemical equations, it becomes quite easy for you to solve them. However, it still maintains a certain level of difficulty at the beginner level. As a result of this, you might find yourself shying away from the equations and procrastinating to the level where you get totally and utterly repulsed by them.

However, there are certain tips that help you during such a stage. When you are beginner, you will be solving quite easy problems compared to those you might see in your Chemistry books. During such times, you will need to keep two essential tips in your mind. These tips will help you to easily balance the equations with ease. These tips are:

• Start Balancing With Single Elements – Attempt at balancing those elements first which occur in the form of a single molecule first. Owing to their single nature, they are easily flexible and their coefficient can be easily changed as and when needed in further steps.
• Balance the Hydrogen and Oxygen Molecules at the End – At the beginning, you will come across a lot of equations involving hydrogen and oxygen molecules. Whenever you encounter these, you should interact with these at the end. This is because hydrogen and oxygen molecules often occur together in both the reactant and product side. Once you are done with balancing other elements, focus on these.

Format for Writing a Balanced Equation

Now that you have balanced the assigned chemical reaction, you might be wondering if there is a format for writing these balanced chemical equations. In actuality, there is not said format that you need to mind for arranging the balanced equation. However, it has also been noticed that people in the field of chemistry often prefer to write solid elements and other compounds first, followed by the gaseous elements and single elements. This often acts as an unwritten rule which is followed by a lot of people around the world.

The Coefficients in a Balanced Chemical Equation

Up until this point of balancing your chemical equations, you might have known about the various facets surrounding the chemical equation. But there is still one significant aspect of balancing which we haven’t discussed: The role of coefficients while balancing equations.

At some point or another, you might have certainly wondered how are these coefficients be used while balancing the equation. After all, we cannot magically create or destroy elements during a chemical reaction. The Law of Conservation of Mass prevents this. In actuality, these coefficients define the ratios. For the reactant side, the coefficients define the ratio in which the substances are being used. And for the product side, the coefficients define the ratio in which the substances are being produced.

What a Balanced Chemical Equation Does Not Tell Us

Balanced chemical equations are highly informative in nature. They divulge a lot of information which is implemented for deriving the desired results from the reactions. However, there are certain aspects which balanced chemical equations don’t make you aware of just by solving the equations. The most prominent aspects amongst these are the subscripts used.

Take, for example, the last chemical equation which we balanced.

Now, if you notice, the element Fe has the subscript 2 beside itself, signifying the number of atoms. But if you notice on the product side, element lacks any subscript. This is quite similar to the oxygen element as well. On one hand, it has the subscript 3 while it has the subscript 2 on the other hand.

In spite of all this, the total mass of the individual atoms present on both sides of the equation is equal to each other. This is due to the Law of Conservation of Mass which ensures that matter isn’t created nor destroyed during a chemical reaction. This is also the reason why the total number of individual atoms are equal on both the reactant and product side.

Rules for Balancing Chemical Equations

By this point, you might have become nicely acquainted with balancing chemical equations on your own. Resulting, the rules associated with balancing chemical equations must also have become imprinted within your mind. It is with the help of such rules that you can easily balance the assigned chemical equations. However, it is equally important that you put these rules on paper and revise them once thoroughly.

Here are a few of the most prominent rules, include:

• Keep The Placement of Reactants and Products in Mind – In every chemical equation, there are two parts to an equation. These parts are separated by an arrow. While writing down the chemical equation, take care to list all the reactants on the left-hand side of the arrow. Similarly, you should take care to list all the products on the right side of the arrow.
• Ensure That The Right Arrow Is Placed – In most cases, the reactants and the products are separated by a single-sided arrow. This signifies a reaction which is irreversible or is unchangeable after a certain stage. However, in certain situations, the reactions occur at equilibrium. This means that reaction at any forward rate results in a reverse reaction. In such situations, the arrow used is two-sided, i.e. facing towards the reactants and the products.
• Emphasize on The Law of Conservation of Mass – While balancing the equations, it is of a predominant nature that you keep applying the Law of Conservation of Mass. This is because matter can neither be produced nor destroyed. Keeping this law in mind greatly helps you while balancing equations. Whenever you find an element which has more or less number of molecules, you can easily place a coefficient to balance it.
• Start With Independent Elements – When you start to balance the equation, start by balancing the independent elements. These are the elements which appear in individually in the equation. If there is no such element or if these elements are already balanced, proceed with the elements that exist in conjunction with other elements. Once this particular element is balanced, you should proceed on to balance other elements until all the elements are balanced.
• Balanced Only With Coefficients – While balancing the chemical equations, balance them only by placing coefficients in front of them. By no means should you add subscripts because this will completely change the formula of the particular reactant or compound, causing a change in the entire meaning that the equation wants to render.

Balancing Chemical Equations with Matrices

Up until this point, you have been balancing chemical equations by the means of trial and error. The process was simple, you had to place a coefficient, check if the other elements were balanced or not, and repeat all the steps until you had all the elements balanced.

However, it won’t be long before you face even tougher balancing problems. And you will face innumerous problems while using the trial and error method for such tough equations. Therefore, on such occasions, you will need a more versatile method for solving the problems.

Fortunately, there is one such methodology for solving chemical equations. This method involves a matrix which you can use to easily solve even the toughest of equations. Here are the steps that you should follow while solving chemical equations:

• Start by placing an alphabet which acts as a variable coefficient for your elements.
• Arrange all the elements in a column matrix format, as per the subscript values.
• Solve each of these matrices and generate the various equations.
• Individually equate all these equations and place the values generated into the other equations that you generated in Step 4.
• Assume a particular number for each of the values, such that, neither of the values that you derive appear in the form of a fraction and use this number to find out the values of the other coefficients.
• Finally, place these values into the initial chemical reaction to derive your balance equation.

Let’s use a simple example to understand this process. Take this chemical equation for instance:

NO + O 2          →         NO 2

Now, this is quite a simple equation. In fact, you might have even figured out how to balance this equation. In spite of this, we will use a simple methodology to help you understand how the entire process works.

Step 1: Start by placing an alphabet which acts as a variable coefficient for your elements

You can use any alphabet as a variable coefficient. For our purpose, we will be using alphabets X, Y, and Z. We will be placing them in this order:

X NO + Y O 2   →         Z NO 2

Step 2: Arrange all the elements in a column matrix format, as per the subscript values.

You should always follow a format for arranging the elements in a column matrix format. First, start by counting the number of atoms present for every individual occurrence of each element. From our first equation, we can derive that:

No. of N atoms = 1 + No. of N atoms = 0 → No. of N atoms = 1

No. of O atoms = 1 + No. of O atoms = 2 → No. of O atoms = 2

According to this format, we will separate the values of each of the elements according to the number of atoms present. Each of these positions will display a value depending on the number of those elements at that particular location. Hence, this is how we will display the values of the elements that are separated into the form of matrices:

X  + Y     →         Z

Notice that value signifying the elements show that each of the elements acquires a particular row. In essence, the nitrogen element acquires the first row while oxygen acquires the second row.

Step 3: Solve each of these matrices and generate the various equations

Once you have the matrices, you need to solve them and generate the required equations for them. The equations that you generate, generally, depending on the number of elements present within the equation. In this case, we have two elements. Therefore, the equations that formed are:

• X + Y0 = Z or X = Z (Equation i)
• X + 2Y = 2Z (Equation ii)

Step 4: Individually equate all these equations and place the values generated into the other equations that you generated in Step 4.

We have already generated the value of the coefficient X in Equation i. The value of X that we have generated is Z. Therefore, it is time that we focused on Equation ii.

X + 2Y = 2Z

According to Equation i, X = Z. Therefore,

• Z + 2Y = 2Z
• 2Y = 2Z – Z
• Y = ½Z             (Equation iii)

Step 5: Assume a particular number for each of the values, such that, neither of the values that you derive appears in the form of a fraction and use this number to find out the values of the other coefficients.

Once we have generated the final equations, it is time that we used them to generate the final values for our coefficients. In order to do this, we need to assume a particular value for each of the variable coefficients, such that the result does not turn out to be a fractional value.

Let’s start by assuming that the value of Z = 1. If Z = 1, then Y = 1/2 (according to the Equation iii). However, we don’t want a fractional value as our result. Therefore, let’s assume that Z = 2. Now that Z = 2, therefore Y = 1. Resultingly, the value of X = 2, since X = Z (as per equation i).

Step 6: Finally, place these values into the initial chemical reaction to derive your balance equation.

The equation which we had at the beginning, was:

According to the results generated by, the value of the variable coefficients stand as per the following:

Let’s place these values into the equation. Upon doing so, we get:

2NO + O 2        →         2NO 2

Therefore, on the reactant side, we have:

• 2 atoms of N
• 4 atoms of O

And on the product side, we have:

• 2 atoms of N.
• 4 atoms of O.

As there you have it again. A perfectly balanced chemical equation solved with the help of matrices.

Balancing Chemical Equations with Odd Number of Atoms on Elements

Another area wherein balancing becomes a tricky affair is during the presence of odd subscripts or atoms of an element. Let us take into consideration, this particular equation:

NH­­­ 3 + O 2         →         NO + H 2 O

The first thing that you will want to do in such cases is to balance those elements which are present in odd numbers on one side but are present in even numbers on the other side of the chemical equation. In this case, we have hydrogen following such a suit. Let us balance this out first.

2NH­­­ 3 + O 2       →         NO + 3H 2 O

Now, we need to balance nitrogen to equate the reaction.

2NH­­­ 3 + O 2       →         2NO + 3H 2 O

At this point, all the elements present in our chemical equations are balanced… except for oxygen. Hence, you will need to find out a coefficient which can effectively help you to balance the oxygen molecule present on the left-hand side of the reaction. On the reactant side, we have 2 oxygen molecules while on the product side, we have 5 oxygen molecules. Therefore, we will need to find a number which, when multiplied by 2, gives us 5 as the answer. Let this number be x.

Therefore, let us place this value into the equation.

2NH­­­ 3 + 5/2O 2 →         2NO + 3H 2 O

Finally, we need to eliminate the fractional part of the equation. Let’s do so by multiplying the entire chemical equation with 2.

4NH­­­ 3 + 5O 2     →         4NO + 6H 2 O

And there you have it. The equation becomes perfectly balanced.

A Few Examples worth Mentioning

Now that you have covered everything that is to be learned about the basics of balancing chemical equations, you should get yourself acquainted with certain worthwhile chemical equations.

• Chemical Reaction for Photosynthesis

6CO 2 + 6H 2 O              →        C 6 H 12 O 6 + 6O 2

• Chemical Reaction for Cellular Respiration

C 6 H 12 O 6 + 6O 2                        →         CO 2 + H 2 O + ATP

• Chemical Reaction for Ammonium Nitrate and Water

NH 4 NO 3 + Water       →         NH 4 + NO 3

• Chemical Reaction for Magnesium and Hydrochloric Acid

Mg + HCl       →         MgCl 2 + H 2

• Chemical Reaction for Lithium and Water

2 Li + 2H 2 O    →         2 LiOH + H 2

• Chemical Reaction for Calcium Carbonate and Hydrochloric Acid

CaCO 3 + HCl →         CaCl 2 + H 2 O

Using Games and Apps to Learn About Balancing Chemical Equations

It has not escaped our sights that a technologically-savvy world such as ours often uses technological means to better understand any newer concepts that they come across. Keeping this factor in mind, we have brought for you two of the greatest means by which you can enhance your skills at balancing chemical equations while simultaneously enjoying it via your smartphones of computers. Here are the means:

• Balancing Chemical Equations – We often encounter situations in which, whatever we do, we fail to solve the equation that has presented itself to us. And let’s be honest, all of us have been there at some point in our lives. It is in such situations that you might find the need for additional help. And this is exactly what Balancing Chemical Equations aims to do. With the help of this application, you will be easily able to balance the toughest of chemical equations. All that you need to do is to enter the unbalanced reactants and products and by clicking a button, the app displays the balanced chemical equation. You can find the Balancing Chemical Equations application on Google Play for free. Here’s a link for the same.

2. The Balancing Equations Game from PHET – Now, an application can only go so far as to keep you engaged. But this is completely contrary to what games can do at keeping you engaged. One of the most engaging games comes from PHET. At their website, you will find the Balancing Equations Game. Upon choosing the option, you are redirected towards a screen for choosing the difficulty of the game. This game is quite interesting. Having tried it out ourselves, we can assure you that not only is it engaging and entertaining, but it is quite informative as well. Therefore, it is one game that you should play if you want to get better at balancing chemical equations and get entertained for a while. Here’s a link to their website.

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• Chemistry Concept Questions and Answers

Balanced Chemical Equations Questions

Chemical equations are symbolic representations of chemical reactions that express the reactants and products in terms of their respective chemical formulae. They also use symbols to represent factors such as reaction direction and the physical states of the reacting entities.

Balanced Chemical Equations Chemistry Questions with Solutions

Q1. A balanced chemical equation is in accordance with-

• Multiple proportion
• Reciprocal proportion
• Conservation of mass
• Definite proportions

Correct Answer: (c) Law of Conservation of Mass

Q2. The correct balanced equation for the reaction __C 2 H 6 O + __O 2 → __CO 2 + __H 2 O is-

• 2C 2 H 6 O + O 2 → CO 2 + H 2 O
• C 2 H 6 O + 3O 2 → 2CO 2 + 3H 2 O
• C 2 H 6 O + 2O 2 → 3CO 2 + 3H 2 O
• 2C 2 H 6 O + O 2 → 2CO 2 + H 2 O

Correct Answer: (b) C 2 H 6 O + 3O 2 → 2CO 2 + 3H 2 O

Q3. The correct balanced equation for the reaction __KNO 3 + __H 2 CO 3 → __K 2 CO 3 + __HNO 3 is-

• 2KNO 3 + H 2 CO 3 → K 2 CO 3 + 2HNO 3
• 2KNO 3 + 2H 2 CO 3 → K 2 CO 3 + 2HNO 3
• KNO 3 + H 2 CO 3 → K 2 CO 3 + 2HNO 3
• 2KNO 3 + 2H 2 CO 3 → K 2 CO 3 + 3HNO 3

Correct Answer: (a) 2KNO 3 + H 2 CO 3 → K 2 CO 3 + 2HNO 3

Q4. The correct balanced equation for the reaction __CaCl 2 + __Na 3 PO 4 → __ Ca 3 (PO 4 ) 2 + __ NaCl is-

• 2CaCl 2 + 2Na 3 PO 4 → 2Ca 3 (PO 4 ) 2 + NaCl
• CaCl 2 + Na 3 PO 4 → Ca 3 (PO 4 ) 2 + NaCl
• 3CaCl 2 + 2Na 3 PO 4 → Ca 3 (PO 4 ) 2 + 6NaCl
• 3CaCl 2 + 2Na 3 PO 4 → Ca 3 (PO 4 ) 2 + 3NaCl

Correct Answer- (c) 3CaCl 2 + 2Na 3 PO 4 → Ca3(PO 4 ) 2 + 6NaCl

Q5. The correct balanced equation for the reaction __TiCl 4 + __H 2 O → __TiO 2 + __HCl is-

• TiCl 4 + 2H 2 O → TiO 2 +2HCl
• TiCl 4 + 2H 2 O → TiO 2 + 4HCl
• 2TiCl 4 + H 2 O → 2TiO 2 + HCl
• TiCl 4 + 4H 2 O → TiO 2 + 4HCl

Correct Answer- (b) TiCl 4 + 2H 2 O → TiO 2 + 4HCl

Q6. Write a balanced equation for the reaction of molecular nitrogen (N 2 ) and oxygen (O 2 ) to form dinitrogen pentoxide.

Answer. The equation for the reaction is-

N 2 + O 2 → N 2 O 5 (unbalanced equation)

The balanced chemical equation is-

2N 2 + 5O 2 → 2N 2 O 5

Q7. On what basis is a chemical equation balanced?

Answer. A chemical equation is balanced using the law of conservation of mass, which states that “matter cannot be created nor destroyed.”

Q8. What is the balanced equation for the reaction of photosynthesis?

Answer. The balanced chemical equation for the reaction of photosynthesis is-

6CO 2 + 6H 2 O → C 6 H 12 O 6 + 6O 2 .

Q9. We must solve a skeletal chemical equation.” Give a reason to justify the statement.

Answer. Skeletal chemical equations are unbalanced. Due to the law of conservation of mass, we must balance the chemical equation. It states that ‘matter cannot be created or destroyed.’ As a result, each chemical reaction must have a balanced chemical equation.

Q10. What does it mean to say an equation is balanced? Why is it important for an equation to be balanced?

Answer. The chemical equation must be balanced in order to obey the law of conservation of mass. A chemical equation is said to be balanced when the number of different atoms of elements in the reactants side equals the number of atoms in the products side. Balancing chemical equations is a trial-and-error process.

Q11. What is meant by the skeletal type chemical equation? What does it represent? Using the equation for electrolytic decomposition of water, differentiate between a skeletal chemical equation and a balanced chemical equation.

Answer. Skeletal equations are those in which formulas are used to indicate the chemicals involved in a chemical reaction.

The law of conservation of mass does not apply to skeletal equations.

The chemical formulas are represented by balanced chemical equations, which follow the law of conservation of mass, which states that the atoms on the reactant and product sides are the same.

H 2 O → H 2 + O 2 : Skeletal equation

2H 2 O → 2H 2 + O 2 : Balanced chemical equation

Q12. Write the balanced chemical equation for the following reaction:

• Phosphorus burns in presence of chlorine to form phosphorus pentachloride.
• Burning of natural gas.
• The process of respiration.
• P 4 + 10Cl 2 → 4PCl 5
• CH 4 +2O 2 → CO 2 +2H 2 O + heat energy
• C 6 H 12 O 6 + 6O 2 + 6H 2 O → 6CO 2 + 12H 2 O + energy

Q13. What Is the Distinction Between a Balanced Equation and a Skeleton Equation?

Answer. The primary distinction between a balanced equation and a skeleton equation is that the balanced equation provides the actual number of molecules of each reactant and product involved in the chemical reaction, whereas a skeleton equation only provides the reactants. Furthermore, a balanced equation may or may not contain stoichiometric coefficients, whereas a skeleton equation does not.

Q14. Balance the equations

• HNO +Ca(OH) 2 → Ca(NO 3 ) 2 + H 2 O
• NaCl + AgNO 3 → AgCl + NaNO 3
• BaCl 2 +H 2 SO 4 → BaSO 4 +HCl

Answer. The balanced chemical equation for the reactions are as follows-

• 2HNO 3 + Ca(OH) 2 → Ca(NO 3 ) 2 + 2H 2 O
• BaCl 2 +H 2 SO 4 → BaSO 4 + 2HCl

Q15. Write a balanced molecular equation describing each of the following chemical reactions.

• Solid calcium carbonate is heated and decomposes to solid calcium oxide and carbon dioxide gas.
• Gaseous butane, C 4 H 10 , reacts with diatomic oxygen gas to yield gaseous carbon dioxide and water vapour.
• Aqueous solutions of magnesium chloride and sodium hydroxide react to produce solid magnesium hydroxide and aqueous sodium chloride.
• Water vapour reacts with sodium metal to produce solid sodium hydroxide and hydrogen gas.
• CaCO 3 → CaO + CO 2

On heating, 1 mol of solid calcium carbonate yields 1 mol of calcium oxide and 1 mol of carbon dioxide gas.

• 2C 4 H 10 +13O 2 → 8CO 2 + 10H 2 O When 2 moles of butane gas react with 13 moles of diatomic oxygen gas, 8 moles of carbon dioxide gas and 10 moles of water vapours are produced.
• MgCl 2 + 2NaOH → 2NaCl + Mg(OH) 2

1 mol magnesium Cordelia reacts with two moles of sodium hydroxide to produce two moles of aqueous sodium chloride solution and one mole of solid magnesium hydroxide.

• 2H 2 O + 2Na → 2NaOH + H 2

2 moles of water vapour react with 2 moles of sodium metal, yielding 2 moles of solid sodium hydroxide and 1 mol of hydrogen gas.

Practise Questions on Balanced Chemical Equations

Balance the following equations-

1. (NH 4 ) 2 Cr 2 O 7 (s) → Cr 2 O 3 (s) + N 2 (g) + H 2 O(g)

2. Ca(OH) 2 + H 3 PO 4 → Ca 3 (PO 4 ) 2 + H 2 O

3. FeCl 3 + NH 4 OH → Fe(OH) 3 + NH 4 Cl

4. Al 2 (CO 3 ) 3 + H 3 PO 4 → AlPO 4 + CO 2 + H 2 O

5. S 8 + F 2 → SF 6

Click the PDF to check the answers for Practice Questions. Download PDF

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Questions on Balancing Chemical Equations - Teachoo Questions - Chapter 1 Class 10 - Chemical Reactions and Equations

Last updated at May 29, 2023 by Teachoo

Write the balanced equation for the following chemical reactions:

Sodium + water → sodium hydroxide + hydrogen.

• 2Na(s) + 2H 2 O(l) → 2NaOH(aq) + H 2 (g)

Na(s) + H 2 O(l) → NaOH(aq) + H 2 (g)

2na(s) + 2h 2 o(l) → naoh(aq) + h 2 (g), na(s) + h 2 o(l) → 2naoh(aq) + h 2 (g).

Explanation - In a balanced chemical equation, the number of atoms of each element is balanced on the left and right side of the equation.

• Na(s) + H 2 O(l) → NaOH(aq) + H 2 (g) [ Unbalanced equation ]
• Na(s) + H 2 O(l) → 2 NaOH(aq) + H 2 (g)
• 2 Na(s) + 2 H 2 O(l) → 2NaOH(aq) + H 2 (g)
• 2Na(s) + 2H 2 O(l) → 2NaOH(aq) + H 2 (g) [ Balanced Equation ]

Balance the equation P 4 O 10 + H 2 O → H 3 PO 4

• P 4 O 10 + H 2 O → H 3 PO 4  [ Unbalanced Equation ]
• P 4 O 10 + H 2 O → 4 H 3 PO 4 
• P 4 O 10 + 6 H 2 O → 4H 3 PO 4 
• P 4 O 10 + 6H 2 O → 4H 3 PO 4   [ Balanced Equation ]

Balance the equation CO 2 + H 2 O → C 6 H 12 O 6 + O 2

• CO 2 + H 2 O → C 6 H 12 O 6 + O 2   [ Unbalanced Equation ]
• 6 CO 2 + H 2 O → C 6 H 12 O 6 + O 2  
• 6CO 2 + 6 H 2 O → C 6 H 12 O 6 + O 2  
• 6CO 2 + 6H 2 O → C 6 H 12 O 6 + 6 O 2  
• 6CO 2 + 6H 2 O → C 6 H 12 O 6 + 6O 2   [ Balanced Equation ]

Balance the equation SiCl 4 + H 2 O → H 4 SiO 4 + HCl

• SiCl 4 + H 2 O → H 4 SiO 4 + HCl [ Unbalanced Equation ]
• SiCl 4 + H 2 O → H 4 SiO 4 + 4 HCl 
• SiCl 4 + 4 H 2 O → H 4 SiO 4 + 4HCl
• SiCl 4 + 4H 2 O → H 4 SiO 4 + 4HCl  [ Balanced Equation ]

Balance the equation Fe + H 2 O → Fe 3 O 4 + H 2

• Fe + H 2 O → Fe 3 O 4 + H 2   [ Unbalanced Equation ]
• 3 Fe + H 2 O → Fe 3 O 4 + H 2
• 3Fe + 4 H 2 O → Fe 3 O 4 + H 2
• 3Fe + 4H 2 O → Fe 3 O 4 + 4 H 2
• 3Fe + 4H 2 O → Fe 3 O 4 + 4H 2  [ Balanced Equation ]

Balance the equation MnO 2 + HCl →  MnCl 2 + Cl 2 + H 2 O

• MnO 2 + HCl →  MnCl 2 + Cl 2 + H 2 O  [ Unbalanced Equation ]
• MnO 2 + 4 HCl →  MnCl 2 + Cl 2 + H 2 O
• MnO 2 + 4HCl →  MnCl 2 + Cl 2 + 2 H 2 O
• MnO 2 + 4HCl →  MnCl 2 + Cl 2 + 2H 2 O  [ Balanced Equation ]

Balance the equation HNO 3 + Ca(OH) 2 → Ca(NO 3 ) 2 + H 2 O

•  HNO 3 + Ca(OH) 2 → Ca(NO 3 ) 2 + H 2 O  [ Unbalanced Equation ]
•   2 HNO 3 + Ca(OH) 2 → Ca(NO 3 ) 2 + H 2 O
•  2HNO 3 + Ca(OH) 2 → Ca(NO 3 ) 2 + 2 H 2 O
•  2HNO 3 + Ca(OH) 2 → Ca(NO 3 ) 2 + 2H 2 O  [ Balanced Equation ]

Balance the equation Al + HCl → AlCl 3 + H 2

• Al + HCl → AlCl 3 + H 2  [ Unbalanced Equation ]
• Al + 3 HCl → AlCl 3 + H 2  
• Al + (3x2) HCl → AlCl 3 + 3 H 2 

Al +    6   HCl → AlCl 3 + 3 H 2 

• Al + 6HCl → 2 AlCl 3 + 3H 2  
• 2 Al + 6HCl → 2AlCl 3 + 3H 2 
• 2Al + 6HCl → 2AlCl 3 + 3H 2  [ Balanced Equation ]

Balance the equation Na 2 CO 3 + HCl → NaCl + H 2 O + CO 2

• Na 2 CO 3 + HCl → NaCl + H 2 O + CO 2   [ Unbalanced Equation ]
• Na 2 CO 3 + HCl → 2 NaCl + H 2 O + CO 2
• Na 2 CO 3 + 2 HCl → 2NaCl + H 2 O + CO 2
• Na 2 CO 3 + 2HCl → 2NaCl + H 2 O + CO 2  [ Balanced Equation ]

Balance the equation C 7 H 6 O 2 + O 2 → CO 2 + H 2 O

• C 7 H 6 O 2 + O 2 → CO 2 + H 2 O  [ Unbalanced Equation ]
• C 7 H 6 O 2 + O 2 → 7 CO 2 + H 2 O
• C 7 H 6 O 2 + O 2 → 7CO 2 + 3 H 2 O

Now, Oxygen needs to be balanced and Oxygen is present in 2 molecules; C 7 H 6 O 2 and O 2 .

For convenience is solving the equation for Oxygen, let us multiply the whole equation by 2 so that the coefficients are even numbers.

• 2 C 7 H 6 O 2 + 2 O 2 → 14 CO 2 + 6 H 2 O
• 2C 7 H 6 O 2 + 15 O 2 → 14CO 2 + 6H 2 O
• 2C 7 H 6 O 2 + 15O 2 → 14CO 2 + 6H 2 O  [ Balanced Equation ]

Balance the equation Fe 2 (SO 4 ) 3 + KOH → K 2 SO 4 + Fe(OH) 3

• Fe 2 (SO 4 ) 3 + KOH → K 2 SO 4 + Fe(OH) 3
• Fe 2 (SO 4 ) 3 + KOH → K 2 SO 4 + 2 Fe(OH) 3
• Fe 2 (SO 4 ) 3 + KOH → 3 K 2 SO 4 + 2Fe(OH) 3
• Fe 2 (SO 4 ) 3 + 6 KOH → 3K 2 SO 4 + 2Fe(OH) 3
• Fe 2 (SO 4 ) 3 + 6KOH → 3K 2 SO 4 + 2Fe(OH) 3  [ Balanced Equation ]

Balance the equation Ca 3 (PO 4 ) 2 + SiO 2 → P 4 O 10 + CaSiO 3

Ca (PO 4 ) 2 + SiO 2 → P 4 O 10 + CaSiO 3

• Ca 3 (PO 4 ) 2 + SiO 2 → P 4 O 10 + 3 CaSiO 3
• 2 Ca 3 (PO 4 ) 2 + SiO 2 → P 4 O 10 + 3CaSiO 3

Now Ca has to be balanced again, so;

• 2Ca 3 (PO 4 ) 2 + SiO 2 → P 4 O 10 + 6 CaSiO 3
• 2Ca 3 (PO 4 ) 2 + 6 SiO 2 → P 4 O 10 + 6CaSiO 3
• 2Ca 3 (PO 4 ) 2 + 6SiO 2 → P 4 O 10 + 6CaSiO 3   [ Balanced Equation ]

Balance the equation KClO 3 → KClO 4 + KCl

• KClO 3 → KClO 4 + KCl

This question requires a slightly different approach.

Here we can see that all atoms are balanced except Oxygen. So we can start with Oxygen by cross multiplying the terms with Oxygen

• 4 KClO 3 → 3 KClO 4 + KCl

Now, since the other elements are also balanced, the equation is balanced. 

• 4KClO 3 → 3KClO 4 + KCl  [ Balanced Equation ]

Balance the equation Al 2 (SO 4 ) 3 + Ca(OH) 2 → Al(OH) 3 + CaSO 4

• Al 2 (SO 4 ) 3 + Ca(OH) 2 → Al(OH) 3 + CaSO 4
• Al 2 (SO 4 ) 3 + Ca(OH) 2 → 2 Al(OH) 3 + CaSO 4
• Al 2 (SO 4 ) 3 + Ca(OH) 2 → 2Al(OH) 3 + 3 CaSO 4
• Al 2 (SO 4 ) 3 + 3 Ca(OH) 2 → 2Al(OH) 3 + 3CaSO 4
• Al 2 (SO 4 ) 3 + 3Ca(OH) 2 → 2Al(OH) 3 + 3CaSO 2    [ Balanced Equation ]

Balance the equation H 2 SO 4 + HI → H 2 S + I 2 + H 2 O

• H 2 SO 4 + HI → H 2 S + I 2 + H 2 O  [ Unbalanced Equation ]
• H 2 SO 4 + 2 HI → H 2 S + I 2 + H 2 O
• H 2 SO 4 + 2HI → H 2 S + I 2 + 4 H 2 O

H needs to be balanced further,

• H 2 SO 4 + 8 HI → H 2 S + 4 I 2 + 4H 2 O
• H 2 SO 4 + 8HI → H 2 S + 4I 2 + 4H 2 O  [ Balanced Equation ]

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 . Complete the balanced chemical equation for the following...

Answer & explanation.

C H 3 ​ N H 3 ​ C l ( a q ) ​ + N a O H ( a q ) ​ → C H 3 ​ N H 2 ( a q ) ​ + H 2 ​ O ( l ) ​ + N a C l ( a q ) ​

Given:  C H 3 ​ N H 3 ​ C l ( a q ) ​ + N a O H ( a q ) ​ → ? ? ? ?

Step 1: Determine the acid and base in the reaction.

C H 3 ​ N H 3 ​ C l ( a q ) ​ Weak acid ​ + ​ N a O H ( a q ) ​ Strong base ​ → ​ ​

Step 2: Write the conjugate base product.

C H 3 ​ N H 3 ​ C l ( a q ) ​ Weak acid ​ + ​ N a O H ( a q ) ​ Strong base ​ → ​ C H 3 ​ N H 2 ( a q ) ​ Conjugate base ​

Step 3: Write the water as a product.

C H 3 ​ N H 3 ​ C l ( a q ) ​ Weak acid ​ + ​ N a O H ( a q ) ​ Strong base ​ → ​ C H 3 ​ N H 2 ( a q ) ​ Conjugate base ​ + H 2 ​ O ( l ) ​

Step 4: Write down the salt product.

C H 3 ​ N H 3 ​ C l ( a q ) ​ Weak acid ​ + ​ N a O H ( a q ) ​ Strong base ​ → ​ C H 3 ​ N H 2 ( a q ) ​ Conjugate base ​ + H 2 ​ O ( l ) ​ + N a C l ( a q ) ​

Step 5: Write the overall balanced acid-base reaction.

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