- Ch.1 - Intro to General Chemistry
- Ch.2 - Atoms & Elements
- Ch.3 - Chemical Reactions
- BONUS: Lab Techniques and Procedures
- BONUS: Mathematical Operations and Functions
- Ch.4 - Chemical Quantities & Aqueous Reactions
- Ch.5 - Gases
- Ch.6 - Thermochemistry
- Ch.7 - Quantum Mechanics
- Ch.8 - Periodic Properties of the Elements
- Ch.9 - Bonding & Molecular Structure
- Ch.10 - Molecular Shapes & Valence Bond Theory
- Ch.11 - Liquids, Solids & Intermolecular Forces
- Ch.12 - Solutions
- Ch.13 - Chemical Kinetics
- Ch.14 - Chemical Equilibrium
- Ch.15 - Acid and Base Equilibrium
- Ch.16 - Aqueous Equilibrium
- Ch. 17 - Chemical Thermodynamics
- Ch.18 - Electrochemistry
- Ch.19 - Nuclear Chemistry
- Ch.20 - Organic Chemistry
- Ch.22 - Chemistry of the Nonmetals
- Ch.23 - Transition Metals and Coordination Compounds

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 chemical equation for the following neutralization reaction producing a soluble salt Phosphoric acid neutralizes a potassium hydroxide solution.

Watch Solution

Write a balanced equation for each reaction:

a. Thermal decomposition of witherite (barium carbonate)

b. Neutralization of stomach acid (HCl) by milk of magnesia (magnesium hydroxide)

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Write a balanced equation for each reaction:

(a) “Slaking” of lime (treatment with water)

(b) Combustion of calcium in air

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Although the alkali metal halides can be prepared directly from the elements, the far less expensive industrial route is treatment of the carbonate or hydroxide with aqueous hydrohalic acid (HX) followed by recrystallization. Balance the reaction between potassium carbonate and aqueous hydriodic acid.

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Complete and balance the following equations:

(a) A saltlike (alkaline earth metal) hydride reacting with water,

(b) Reduction of a metal halide by hydrogen to form a metal,

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Complete and balance the following equations:

(a) An active metal reacting with acid,

(b) A saltlike (alkali metal) hydride reacting with water,

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Balance the following equations.

a. ___ C_{3}H_{8}O(l) + ____ O_{2}(g) → ________ CO_{2}(g) + ________ H_{2}O(l)

b. _____ PH_{3}(g) + ____ O_{2}(g) → _______ P_{4}O_{10}(s) + _______ H_{2}O(g)

c. ____ B_{2}O_{3}(s) + _____ HF(l) → _______ BF_{3}(g) + _________ H_{2}O(l)

d) ____Cu_{2}O(s) + ______O_{2}(g) → ______CuO(s)

e) ___CaCl_{2}(aq) + ____K_{2}CO_{3}(aq) → ___ KCl(aq) + ____CaCO_{3}(aq)

f) ___ MgO(s) + ___ Fe(s) → ___ Fe_{2}O_{3}(s) + ___ Mg(s)

Watch Solution

Balance the following equations by inserting coefficients as needed.

_{— }Li_{3}N + _{— }H_{2}O → _{— }LiOH + _{— }NH _{3}

_{— }ZnO + _{— }CO → _{— }Zn + _{— }CO_{2}

_{— }SO_{2} + _{— }O_{2} → _{— }SO_{3}

Watch Solution

Balance the following equations by inserting coefficients as needed.

a. CaCO_{3} + HCl → CaCl_{2} + CO_{2} + H_{2}O

b. C_{6}H_{12}O_{2} + O_{2} → CO_{2} + H_{2}O

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Balance the equation.

HC_{2}H_{3}O_{2}(aq) + Ba(OH)_{2}(aq) → H_{2}O(l) + Ba(C_{2}H_{3}O_{2})_{2}(aq)

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Balance the following equation:

K_{2}CrO_{4} + Na_{2}SO_{3} + HCl → KCl + Na _{2}SO_{4} + CrCl_{3} + H_{2}O

Watch Solution

Balance the following equations by inserting coefficients as needed.

a. _{— }C_{3}H_{8} + _{— }O_{2} → _{— }CO_{2} + _{— }H_{2}O

b. _{— }P_{4}O_{10} + _{— }H_{2}O → _{— }H_{3}PO_{4}

Watch Solution

Complete and balance each of the following equations for acid-base reactions.

a. H_{3}PO_{4}(aq) + Sr(OH)_{2}(aq) →

b. HC_{2}H_{3}O_{2}(aq) + Ca(OH)_{2}(aq) →

c. HCl(aq) + Ba(OH)_{2}(aq) →

Watch Solution

Balance the following equation:

a. Zn +H_{2}SO_{4} → ZnSO_{4} + H_{2}

b. Al + H_{2}SO_{4} → Al_{2}(SO_{4})_{3} + H_{2}

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What kind of reaction occurs when you mix aqueous solutions of barium sulfide and sulfuric acid?

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Complete and balance the following reaction:

H_{3}PO_{4}(aq) + KOH(aq) →

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Calcium hydride reacts with water to form calcium hydroxide (aqueous) and hydrogen gas.

Write a balanced chemical equation for the reaction.

Express your answer as a balanced chemical equation. Identify all of the phases in your answer.

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Balance the following equation:

K_{2}CrO_{4} + Na _{2}SO_{3} + HCl → KCl + Na _{2}SO_{4} + CrCl_{3} + H_{2}O

Enter the coefficients for each compound seperated by commas, in the order in which they appear in the equation.

Watch Solution

Balance the following equations and indicate the type of reaction taking place:

i) ____NaBr + ____H_{3}PO_{4} → ___Na_{3}PO_{4} + ___HBr

ii) ___Ca(OH)_{2} + ___Al_{2}(SO_{4})_{3} → ___CaSO_{4} + __Al(OH)_{3}

Watch Solution

Determine the balanced chemical equation for this reaction.

C_{8}H_{18}(g) + O_{2}(g) → CO_{2}(g) + H_{2}O(g)

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Balance the following equation in acidic conditions. Phases are optional.

H_{3}AsO_{3}+ I_{2} → H_{3}AsO_{4} + I^{-}

(Redox and Oxidation Reactions)

Watch Solution

When iron rusts, solid iron reacts with gaseous oxygen to form solid iron (III) oxide. Write a balanced chemical equation for the reaction. Express as a chemical equation, identify all the phases in the answer.

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What is the coefficient of O _{2} when the following equation is properly balanced with the smallest set of whole numbers?

C_{3}H_{7}OH(I) + O_{2}(g) → CO_{2} + H_{2}O

A) 5

B) 17

C) 9

D) 6

E) 2

Watch Solution

When the chemical reaction representing the complete combustion of acetic acid, C_{2}H_{4}O_{2}, is balanced with the smallest integer coefficients, the sum of the coefficients in the balanced equation is:

A. 9

B. 7

C. 2

D. 4

E. 5

Watch Solution

When the chemical equation

FeS + O_{2} → FeO + SO_{2}

is balanced with the smallest integral coefficients, the coefficient of O _{2} is:

a. 3

b. 2

c. 4

d. 8

e. 5

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When the chemical equation below is balanced with the smallest INTEGER coefficients, the sum of the coefficients is:

(H_{3}C)_{2}PCH_{2}N(CH_{3})_{2} + O_{2} → CO_{2 }+ H_{3}PO_{4} + N_{2} + H_{2}O

a. 13

b. 91

c. 35

d. 41

e. 33

Watch Solution

When the equation below is balanced with the smallest set of coefficients, what is the coefficient of Zn?

Zn + HNO_{3} → Zn(NO_{3})_{2} + NH_{4}NO_{3} + H_{2}O

a. 1 (no coefficient written)

b. 2

c. 3

d. 4

e. 6

Watch Solution

Balance this equation using the smallest possible integers, NO + NH_{3} → N_{2} + H_{2}O. What is the coefficient of the water?

(A) 1

(B) 2

(C) 3

(D) 6

(E) 12

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Balance this equation using the smallest possible integers, S + HNO_{3} → H_{2}SO_{4} + NO_{2} + H_{2}O. What is the coefficient of the water?

(A) 1

(B) 2

(C) 4

(D) 6

(E) 10

Watch Solution

After the following equation is balanced with the smallest possible whole numbers, what is the sum of the coefficients?

NO_{2}(g) + H_{2}(g) → NH_{3}(g) + H_{2}O (l)

(A) 4

(B) 10

(C) 15

(D) 17

(E) 20

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Dinitrogen pentoxide is a gas that decomposes to form nitrogen dioxide gas and oxygen gas. Write a balanced equation for the reaction, and be sure to include the states of matter.

Watch Solution

What is the sum of the coefficients when the following equation is balanced with the smallest whole number coefficients?

C_{12}H_{10}O + O_{2} → CO_{2} + H_{2}O

(A) 31

(B) 32

(C) 64

(D) 65

(E) None of the above

Watch Solution

Balance the following equation and select the correct coefficients for NO_{2} and N_{2}, respectively:

__ NH_{3} _{(g)} + __ NO_{2} _{(g)} → __ N_{2} (g) + __ H_{2}O _{(l)}

a) 3, 5

b) 5, 9

c) 4, 3

d) 6, 7

e) 7, 9

Watch Solution

A reaction occurs between calcium nitrate and ammonium fluoride, producing calcium fluoride, dinitrogen monoxide, and water vapor. The correct set of stoichiometric coefficients, respectively, for the balanced reaction is:

A) 3 6 6 3 4

B) 3 2 1 2 2

C) 1 2 2 1 1

D) 1 2 1 2 4

E) 2 1 2 2 2

Watch Solution

After balancing the following reaction, what are the coefficients for H _{2}O and NaAu(CN)_{2}, respectively?

__ Au + __ NaCN + __ O_{2} + __ H_{2}O → __ NaAu(CN)_{2} + __ NaOH

A) 1, 3

B) 2, 4

C) 1, 4

D) 3, 2

E) 3, 1

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

Watch Solution

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)

Watch Solution

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

Watch Solution

When hydrogen peroxide (H_{2}O_{2}) 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

Watch Solution

What is the stoichiometric coefficient for oxygen gas (O_{2}) when the following equation is balanced using the lowest, whole-number coefficients?

_____C_{3}H_{8}(*g*) + _____ O_{2}(*g*) → _____ CO_{2}(*g*) + _____ H_{2}O(*l*)

A) 9 B) 7 C) 5 D) 3 E) 1

Watch Solution

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

Watch Solution

What is the coefficient of H_{2}O when the following equation is properly balanced with the smallest set of whole numbers?

___ Al_{4}C_{3} + ___ H_{2}O → ___ Al(OH)_{3} + ___ CH_{4}

A) 3

B) 4

C) 6

D) 12

E) 24

Watch Solution

What is the coefficient of H_{2}O when the following equation is properly balanced with the smallest set of whole numbers?

___ Na + ___ H_{2}O → ___ NaOH + ___ H_{2}

A) 1

B) 2

C) 3

D) 4

E) 5

Watch Solution

Balance the following equation: B_{2}O_{3}(s) + HF(l) → BF_{3}(g) + H_{2}O(l)

A) B_{2}O_{3} (s) + 6HF(l) → 2BF_{3} (g) + 3H_{2}O(l)

B) B_{2}O_{3} (s) + H_{6}F_{6}(l) → B2F_{6}(g) + H_{6}O_{3} (l)

C) B_{2}O_{3} (s) + 2HF(l) → 2BF_{3} (g) + H_{2}O(l)

D) B_{2}O_{3} (s) + 3HF(l) → 2BF_{3} (g) + 3H_{2}O(l)

E) B_{2}O_{3} (s) + 6HF(l) → 2BF_{3} (g) + 6H_{2}O(l)

Watch Solution

Write the balanced chemical equation:

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

Watch Solution

Balance the following chemical equation.

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

Watch Solution

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

Watch Solution

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

Watch Solution

Write a balanced equation describing the neutralization of sodium hydroxide with sulfuric acid. In the balanced equation, what is the sum of the coefficients.

A. 6

B. 4

C. 5

D. 7

E. 8

Watch Solution

Balance the following chemical equations:

Watch Solution

Write the balanced chemical formula for the reaction in which the reactant potassium chlorate is broken down into two products: potassium chloride and oxygen.

Watch Solution

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

Watch Solution

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

Watch Solution

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)

Watch Solution

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)

Watch Solution

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

Watch Solution

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

Watch Solution

**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

Watch Solution

The balancing of chemical equations is based on the Law of Conservation of Mass, which states that matter cannot be created nor destroyed. This means that the number of reactant atoms must be the same as the number of product atoms.

We call these elements, which don't change after a chemical reaction, *immutable groups*.

**Intro: Why Must We Balance Chemical Equations? **

A chemical equation serves as a symbolic representation of a chemical reaction and so knowing the numbers of each compounds involved is essential.

For the following equation the numbers in **blue** are called your coefficients and represent the number of moles for each compound shown.

This written chemical reaction tells us a story that states, “2 moles of hydrogen, H_{2 }, react with 1 mole of oxygen , O_{2} , to produce 2 moles of water, H_{2}O. ”

The numbers in **red** are the subscripts and combined with the coefficients give us the number of each particular element.

Determining the number of each element will also be needed for later calculations dealing with reaction stoichiometry, limiting reagent and identifying reaction types.

**Instructions for Balancing**

Balancing a chemical reaction doesn’t really require algebra, an Algebraic Balance or even a calculator. All you need to do is follow these instructions and you’ll be balancing like a pro.

**STEP 1: **Set up a list for the elements that are reactants and another list for the elements that are products. If your numerical values in both lists don’t match then you’ll have to balance the chemical equation.

The following chemical reaction deals with the combustion of gaseous butane:

**STEP 2:** Start at the top and going down both lists make sure to balance each element one by one.

**STEP 3:** At this point notice there are 4 carbon atoms as reactants versus 1 carbon atom as a product.

Balance the number of carbon atoms by placing a “4” in front of CO_{2}. The “4” is distributed so the number of carbon and oxygen atoms are both affected.

**STEP 4: **Continuing down the lists we next balance the number of hydrogen atoms.

The reactant side has 10 hydrogen atoms, while the product side has only 2 hydrogen atoms. So placing a 5 in front of H_{2}O will give 10 hydrogen atoms (2 x 5).

In addition the 5 in front of H_{2}O is distributed to oxygen. We now have 5 oxygen atoms from H_{2}O and 8 oxygen atoms from CO_{2}, giving us a current total of 13 oxygen atoms.

**STEP 5: **At this point we need to balance our oxygen atoms and in order to have 13 oxygen atoms on both sides we place “6.5” in front of O_{2}.

**STEP 6:** Although the chemical equation is balanced because both lists have identical numerical values, you want your coefficients to be whole numbers and not decimals or fractions.

**STEP 7:** This additional step is only necessary when you have a decimal or fraction as a coefficient. When balancing a chemical equation you need all the coefficients to be whole numbers.

In this example, in order to convert 6.5 into a whole number we multiply the entire chemical equation by 2.

**FOR EXPERTS ONLY: **Eventually if you keep up with your chemistry studies you’ll be asked to do more advanced chemical reactions (redox, acids and bases, precipitation) while using simple stoichiometry.

You may have to balance redox reactions in acidic solutions or basic solutions. Again you won’t have to balance these reactions algebraically or with advanced math.

You may even be given just the reactants without the products. In this case you’ll have to deal with charges that are written as superscripts. These charged ions will be in the form of a cation and an anion and will include finding the net ionic equation.

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