## Chem – Balancing Chemical Equations

How do you balance a chemical equation?

The next step is to balance entire chemical equations. Balancing chemical equations comes from the concept that the amount of stuff you put into the reactants side of the chemical equation should be the same amount of stuff you get out of the products side of the chemical equation. Another way that chemistry books and teachers usually say it is that mass is not gained or lost in a chemical equation. It is like baking cookies. If you double the ingredients you mix together then you get double the cookies in the end.

Remember to always total up all the elements on both sides of the chemical equation and any chemical compound not showing a number in front of it has a coefficient of 1. Don’t try to total them in your head. Write them down on a piece of paper or drawing program on your computer. It takes approximately 15 attempts at balancing chemical equations until you get comfortable with it. It takes approximately 30 attempts until you are good at it. It takes approximately 60 attempts until you can actually start do these in your head. If you do not write down the first 60 attempts on paper you will never be able to do them in your head. Trust me I have done thousands of them. The strategy is to take it slow and steady with balancing. Don’t get frustrated; just keep trying to move forward.

Examples: Balance the following chemical equations.

 ___N2(g) + ___O2(g) —-> ___N2O3(g) 2 N2(g) + 3 O2(g) —-> 2 N2O3(g) 1 ___CaCl2(aq) + ___Al2(SO3)3(aq) —-> ___CaSO3(aq) + ___AlCl3(aq) 3 CaCl2(aq) + Al2(SO3)3(aq) —-> 3 CaSO3(aq) + 2 AlCl3(aq)

VIDEO Balancing Chemical Equation Demonstrated Example 1: Balance the unbalanced chemical equation below.

MgBr2(aq) + NaI(aq) ——> MgI2(aq) + NaBr(aq)

Step 1:

How do we start to balance a chemical equation?

Answer: First we write down the chemical equations on our paper and then we draw a line down the yield sign to separate the two halves of a chemical equation.

 MgBr2(aq) + NaI(aq) — —> MgI2(aq) + NaBr(aq) Mg = Mg = Br = Br = Na = Na = I = I =

Step 2:

What do we do next?

Answer: Below the reactants and products side we list each element followed by an equal sign. Make sure you write both sides in the same element order to save you head ache later.

 MgBr2(aq) + NaI(aq) — —> MgI2(aq) + NaBr(aq) Mg = Mg = Br = Br = Na = Na = I = I =

Step 3:

Now count up all the elements on each side of the equation and assume that all the coefficients are 1.

 MgBr2(aq) + NaI(aq) — —> MgI2(aq) + NaBr(aq) Mg = 1 Mg = 1 Br = 2 Br = 1 Na = 1 Na = 1 I = 1 I = 2

Step 4:

What we want to focus on from here on out is the differences in the numbers between the reactants and products. What differences do you notice in this problem?

 MgBr2(aq) + NaI(aq) — —> MgI2(aq) + NaBr(aq) Mg = 1 Mg = 1 Br = 2 Br = 1 Na = 1 Na = 1 I = 1 I = 2

Step 5:

We can only change these differences by using coefficients to change the chemical equation. We CANNOT EVER change the subscripts. You can start by changing any coefficient you like. It does not really matter where you start. I will begin by changing the coefficient of NaBr to a 2.

 MgBr2(aq) + NaI(aq) — —> MgI2(aq) + 2 NaBr(aq) Mg = 1 Mg = 1 Br = 2 Br = 1 Na = 1 Na = 1 I = 1 I = 2

Step 6:

Immediately after that you want to change the number of each element below the side you changed. The 2 NaBr changes the Br but it also affects the Na.

 MgBr2(aq) + NaI(aq) — —> MgI2(aq) + 2 NaBr(aq) Mg = 1 Mg = 1 Br = 2 Br = 2 Na = 1 Na = 2 I = 1 I = 2

Step 7:

Now where is there a difference?

Answer: The Na and I. So I will put a 2 coefficient in front of NaI.

 MgBr2(aq) + 2 NaI(aq) — —> MgI2(aq) + 2 NaBr(aq) Mg = 1 Mg = 1 Br = 2 Br = 2 Na = 2 Na = 2 I = 2 I = 2

Step 8:

As we check for any more differences we find none. Therefore, we are done balancing the equation.

COMPLETE ANSWER: MgBr2(aq) + 2 NaI(aq) ——> MgI2(aq) + 2 NaBr(aq)

VIDEO Balancing Chemical Equation Demonstrated Example 2:Please balance the unbalanced chemical equation below.

N2(g) + H2(g) —-> NH3(g)

Step 1:

How do we start to balance a chemical equation?

Answer: First we write down the chemical equations on our paper and then we draw a line down the yield sign to separate the two halves of a chemical equation.

 N2(g) + H2(g) — –> NH3(g) N = N = H = H =

Step 2:

What do we do next?

Answer: Below the reactants and products side we list each element followed by an equal sign. Make sure you write both sides in the same element order to save you head ache later.

 N2(g) + H2(g) — –> NH3(g) N = N = H = H =

Step 3:

Now count up all the elements on each side of the equation and assume that all the coefficients are 1.

 N2(g) + H2(g) — –> NH3(g) N = 2 N = 1 H = 2 H = 3

Step 4:

What we want to focus on from here on out is the differences in the numbers between the reactants and products. What differences do you notice in this problem?

 N2(g) + H2(g) — –> NH3(g) N = 2 N = 1 H = 2 H = 3

Step 5:

We can only change these differences by using coefficients to change the chemical equation. We CANNOT EVER change the subscripts. You can start by changing any coefficient you like. It does not really matter where you start. I will begin by changing the coefficient of NH3 to a 2.

 N2(g) + H2(g) — –> 2 NH3(g) N = 2 N = 1 H = 2 H = 3

Step 6:

Immediately after that you want to change the number of each element below the side you changed. The 2 NH3 changes the N but it also affects the H.

 N2(g) + H2(g) — –> 2 NH3(g) N = 2 N = 2 H = 2 H = 6

Step 7:

Now where is there a difference?

Answer: The H. So I will put a 3 coefficient in front of the H2.

 N2(g) + 3 H2(g) — –> 2 NH3(g) N = 2 N = 2 H = 6 H = 6

Step 8:

As we check for any more differences we find none. Therefore, we are done balancing the equation.

COMPLETE ANSWER: N2(g) + 3 H2(g) —-> 2 NH3(g)

PRACTICE PROBLEMS: Balance the chemical equations below.

 H2O(l) —> H2(g) + O2(g) Answer: 2 H2O(l) —> 2 H2(g) + O2(g) H2SO4(aq) —> H2(g) + S(s) + O2(g) Answer: H2SO4(aq) —> H2(g) + S(s) + 2 O2(g) OF2(g) + NH3(g) —> N2F4(g) + O2(g) + H2(g) Answer: 2 OF2(g) + 2 NH3(g) —> N2F4(g) + O2(g) + 3 H2(g) Fe3+(aq) + CO32-(aq) —-> Fe2(CO3)3(s) Answer: 2 Fe3+(aq) + 3 CO32-(aq) —> Fe2(CO3)3(s) Ba(OH)2(aq) + HCl(aq) —-> H2O(l) + BaCl2(aq) Answer: Ba(OH)2(aq) + 2 HCl(aq) —> 2 H2O(l) + BaCl2(aq)  