When balancing an equation always make sure the **types **and **number** of atoms on both sides of the arrow are equal.

Balancing a chemical reaction involves following the Law of Conservation of Mass and Energy. The total number of elements must be the same on both sides.

**Concept:** Balancing Chemical Reactions

Welcome back, guys. In this new video, we're going to handle a topic that's essential for any type of calculations we're going to do this semester. It can be quite easy as long as we follow basic rules. It's all about balancing chemical equations.

We're going to say when balancing an equation, always make sure that the number and types of atoms on both sides of the arrow are equal.

For this example, it says write balanced equations for each of the following by inserting the correct coefficient in the blanks. Coefficient is just the number that goes in front of each one of these compounds in a balanced equation.

The approach we should always take is let's just write a list. Both sides of the equation have the same elements, and for them to be balanced, they have to have the same exact numbers of each of these types of elements. So on the left side, we have only 1 aluminum. We also have, there's a 2 here, so we have 2 chlorines. On the right, we still have 1 aluminum, and we have now 3 chlorines.

You can see that the lists don't match. That's because the equation is not balanced. All we have to do is just go down the list and see what doesn't match. Our chlorines aren't matching, so that's what we want to balance. We have to think what number can we put in front of both of those compounds to make them both have the same number of chlorines?

Here we're going to say lowest common multiple is 6 between 3 and 2, so if I put a 3 in front of that Cl2, it becomes distributed. It's 3 times 2, so this becomes 6. And here I'm going to put a 2. We're going to say 2 times the 3 gives a 6 here also. Our chlorines now match.

But that 2 is also in front of the aluminum, so that's going to get multiplied by 2 as well, so this becomes 2. Our lists are almost equal; we just have to do one more step. All we have to do now is just throw a 2 in front of this aluminum, making this 2. Now both our lists match; now they're balanced.

For the next one, let's take a look to see what groups we have. What we should notice here is, do you see this PO4 right here? It's right here as well. It's a polyatomic ion; it came from phosphate. If you have polyatomic ions that look the same on both sides, just keep them all together to help us balance.

So we're going to say on the left side, we have Ba, we have PO4 – we're keeping it all together, counting it as the same – we also have K, and look, you see this OH right here? That OH is right over here as well. So we're going to have OH in this as well. We'll also have the same exact elements and compounds on the right side.

Now all we have to do is just to see which ones match up and which ones are different. On the left side, we have a 3 here with the Ba, with barium, so we have 3 bariums. We have a 2 on the outside, which is going to mean we have 2 phosphates. We have 1 potassium right here, and 1 OH. On the right side, we only have 1 Ba, we only have 1 phosphate, we have a little 3 here, so we have 3 potassiums, and then we have a 2 out here, so we have 2 hydroxide ions.

Now, let's go down the list and match things up. For barium, we have 3 on the left but 1 on the right, so let's put a 3 here. That 3 will get distributed; this becomes 3, and then 3 times the 2 out here gives us 6 OH. For phosphate, we have 2 on the left but only 1 on the right, so we're going to put a 2 here, which makes this 2. But then it gets distributed with this 3, so 3 times 2 gives us 6 potassium. All we have to do now is our potassiums don't match, so I'm going to put a 6 here. That's going to get distributed, making the potassium 6 and also my OH 6.

Both our lists match up, so now it's balanced. If you want, you could put a 1 here to show that there's a 1 there. You don't really have to, but if you wanted to, you could just put it there.

This last one. This one is an example of combustion. We have a carbon and hydrogen reacting with O2 to give us CO2 and water. I'm going to minus myself from the shot, guys, so that we have more space to work with.

Like before, we're just going to make a list. Make sure things match up. So we have carbon, hydrogen, and oxygen here; carbon, hydrogen, and oxygen here. What we're going to do now is we're going to say we have 4 carbons, 10 hydrogens, and then we have 2 oxygens. On the other side we have 1 carbon, 2 hydrogens; we have 2 oxygens here and 1 oxygen over here, so this is 3.

Now we just have to make sure things balance, so we have 4 carbons on the left, only 1 on the right, so we're going to throw a 4 here. This changes this to 4, but then 4 times this 2 gives me 8 oxygens, plus the 1 oxygen we have here gives me 9.

Next we want to look at the hydrogens. On the left we have 10, but on the right we only have 2, so we're going to throw a 5 here; 5 gets multiplied with the 2 to give me 10 hydrogens. Then we have 4 times this 2 here gives me 8 oxygens here, and then 5 times 1 gives me 5 oxygens here. So 8 oxygens here, 5 oxygens here gives me 13 oxygens total.

Now what we have to do here is we need to balance this out further. The only number I can put right here to give me 13 is if I put 6.5, because 6.5 times 2, the 2 right here, gives me 13 as well. With combustion analysis, this happens sometimes. This would not be the best answer, because we can't have decimals or fractions as coefficients. It has to be whole numbers. So what we're going to do here is we're just going to multiply this entire thing times 2. There's a 1 here.

So all the coefficients inside the brackets will get multiplied by 2, and that's going to give me a new equation, 2 C4H10 plus the 6.5 gets multiplied by the 2 on the outside, so that's 13 O2. Gives me 8 CO2 plus 10 H2O. Just like that, guys, we've balanced everything.

Balancing can be really easy, as long as you make your list and follow where the numbers go. And remember, balancing may seem very simple, but it's essential for any type of calculations we're going to do later on, because later on, we're going to be tackling chemical calculations where we need a balanced equation in order to do them.

So guys, just keep practicing, look at different types of balancing problems, and it can be really easy as long as you implement this type of system.

Write a balanced equation to show the reaction of sulfurous acid with lithium hydroxide to form water and lithium sulfite.

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Write a balanced chemical equation for the reaction of aqueous solutions of magnesium chloride and potassium phosphate.

a. MgCl_{2}(aq) + K_{3}PO_{4}(aq) → K_{3}Mg(s) + PO_{4}Cl_{2}(aq)

b. 3 MgCl_{2}(aq) + 2 K_{3}PO_{4}(aq) → 3 K_{2}Mg(s) + 2 PO_{4}Cl_{3}(aq)

c. MgCl(aq) + KPO_{4}(aq) → MgPO_{4}(s) + KCl(aq)

d. MgCl_{2}(aq) + 2 KPO_{4}(aq) → Mg(PO_{4})_{2}(s) + 2 KCl(aq)

e. 3 MgCl_{2}(aq) + 2 K_{3}PO_{4}(aq) → Mg_{3}(PO_{4})_{2}(s) + 6 KCl(aq)

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When the chemical equation for the complete combustion of toluic acid, CH_{3}C_{6}H_{4}CO_{2}H, to form carbon dioxide and water is balanced, the coefficient in front of the O _{2} is

a. 4

b. 10

c. 9

d. 5

e. 20

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When hydrogen peroxide (H2O2) decomposes, it forms oxygen and hydrogen gas:

____ H_{2}O_{2} → _____ O_{2 }+ _____ H_{2}O

When this equation is balanced, what is the coefficient for oxygen?

a. 1

b. 2

c. 3

d. 4

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When the following equation is balanced, the coefficient of sulfur dioxide is _______.

PbS (s) + O_{2} (g) → PbO (s) + SO_{2} (g)

a. 5

b. 1

c. 3

d. 2

e. 4

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Write the balanced chemical equations.

Aqueous hydrochloric acid reacts with solid manganese(IV) oxide to form aqueous manganese(II) chloride, liquid water, and chlorine gas.

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Balance the following chemical equation.

PCl_{3}(l) + H_{2}O(l) → H_{3}PO_{3}(aq) + HCl(aq)

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What are the coefficients for the decomposition of nitroglycerin?

__ C_{3}H_{5}N_{3}O_{9} → __ N_{2} + __ CO_{2} + __ H_{2}O + __ O_{2}

A. 2,3,6,2,1

B. 2,3,6,5,1

C. 4,6,12,10,12

D. 4,3,12,10,1

E. 4,6,12,10,1

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Balance the following chemical equations involving boron and/or fluorine:

1. B + Br_{2} → BBr_{3}

2. B_{2}O_{3} + C → B_{4}C + CO

3. F_{2} + H_{2}O → O_{2} + HF

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Balance the following chemical equations:

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Write the balanced equation for the following reaction: aqueous sodium sulfide reacts with aqueous iron (III) nitrate to form aqueous sodium nitrate and solid iron (III) sulfide. What are the coefficients in the balanced reaction (write the substances in the order given)?

A. 1, 1, 1, 3

B. 1, 2, 2, 3

C. 3, 2, 6, 1

D. 2, 3, 1, 6

E. 3, 1, 1, 3

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When the following equation is balanced, the coefficient of sulfur dioxide is _____.

PbS (s) + O_{2} (g) → PbO (s) + SO_{2} (g)

A. 5

B. 1

C. 3

D. 2

E. 4

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What is the coefficient of aluminum phosphate once the reaction below is balanced?

______ Hg_{3}(PO_{4})_{2} (aq) + ______ Al (s) → ______ AlPO _{4} (s) + ______ Hg (l)

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What is the sum of the coefficients for the products in the following chemical equation?

______ C_{12}H_{26} (g) + ______ O_{2} (g) → ______ CO_{2} (g) + ______ H_{2}O (l)

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What is the sum of the coefficients for the reactants in the balanced equation for the combustion of

isopropyl alcohol:

______C_{3}H_{8}O + ______O_{2 → }______CO_{2 }+ ______H_{2}O

a) 25

b) 13

c) 11

d) 6

e) none of the given answers

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What is the coefficient of oxygen in the balanced combustion reaction of one (1) mole of acetone (C_{3}H_{6}O)?

a. 1

b. 2

c. 3

d. 4

e. 5

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**For this question, consider what might happen if a solution of mercury (II) nitrate were to be mixed with a solution of ammonium iodide. You will want to write a balanced equation and a net ionic equation to help you answer the following questions.**

In the balanced equation, the coefficient in front of the mercury compound on the left side of the equation and the ammonium compound on the right side of the equation will be

a) 1 and 1, respectively

b) 1 and 2, respectively

c) 2 and 1, respectively

d) 2 and 4, respectively

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