Scientific Tutor

What are some of the tricks for balancing a chemical equation?

For the simple chemical equations in the previous section, the process is pretty straightforward, but there are some chemicals equations that many people have trouble balancing. Those chemical equations usually include situations where you need to keep balancing until you get to high coefficient numbers or when you have lots of elements on both sides of the chemical equations. In this section I will show you some tricks to make the these kinds of chemical equations much easier.

 

VIDEO Balancing Chemical Equation Demonstrated Example 3: Balance the chemical equation below.

 

Al(OH)_3(aq) + Li₂CO_3(aq) ——> LiOH_(aq) + Al₂(CO₃)_3(s)

 

Step 1:

How do we start to balance a chemical equation?

Answer: Same as before, 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.

Al(OH)3(aq) + Li2CO3(aq) —	—> LiOH(aq) + Al2(CO3)3(s)
Al = 1	Al = 2
OH = 3	OH = 1
Li = 2	Li = 1
CO3 = 1	CO3 = 3

 

Step 2:

Then we count up all the elements on each side of the equation and assume that all the coefficients are 1. However, in this case we have polyatimic ions. If the polyatomic ions appear the same on both sides of the chemical equations, we can count the polyatomic ions as one whole instead of individual elements. This makes counting much simpler.

Al(OH)3(aq) + Li2CO3(aq) —	—> LiOH(aq) + Al2(CO3)3(s)
Al = 1	Al = 2
OH = 3	OH = 1
Li = 2	Li = 1
CO3 = 1	CO3 = 3

 

Step 3:

Now start balancing the equation by adding coefficients. I will start with the Aluminum on the left side.

2 Al(OH)3(aq) + Li2CO3(aq) —	—> LiOH(aq) + Al2(CO3)3(s)
Al = 2	Al = 2
OH = 6	OH = 1
Li = 2	Li = 1
CO3 = 1	CO3 = 3

 

Step 4:

Next I try the CO₃ on the left side.

2 Al(OH)3(aq) + 3 Li2CO3(aq) —	—> LiOH(aq) + Al2(CO3)3(s)
Al = 2	Al = 2
OH =6	OH = 1
Li = 6	Li = 1
CO3 = 3	CO3 = 3

 

Step 5:

Then I Li and OH on the right side.

2 Al(OH)3(aq) + 3 Li2CO3(aq) —	—> 6 LiOH(aq) + Al2(CO3)3(s)
Al = 2	Al = 2
OH =6	OH = 6
Li = 6	Li = 6
CO3 = 3	CO3 = 3

 

Step 6:

All the elements and polyatomic ions are the same on both sides so I am done.

COMPLETE ANSWER: 2 Al(OH)_3(aq) + 3 Li₂CO_3(aq) ——> 6 LiOH_(aq) + Al₂(CO₃)_3(s)

 

VIDEO Balancing Chemical Equation Demonstrated Example 4: Balance the chemical equation below.

 

C₅H₇OH_(l) + O_2(g) ——> CO_2(g) + H₂O_(g)

 

Step 1:

How do we start to balance a chemical equation?

Answer: Same as before, 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.

C5H7OH(l) + O2(g) —	—> CO2(g) + H2O(g)
C = 5	C = 1
H = 8	H = 2
O = 3	O = 3

 

Step 2:

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

C5H7OH(l) + O2(g) —	—> CO2(g) + H2O(g)
C = 5	C = 1
H = 8	H = 2
O = 3	O = 3

 

Step 3:

Start to try to balance them with coefficients we are going to start with 5 in front of the CO₂ on the right side. It is best to balance the carbons and hydrogen first in these kinds of reactions.

C5H7OH(l) + O2(g) —	—> 5 CO2(g) + H2O(g)
C = 5	C = 5
H = 8	H = 2
O = 3	O = 11

 

Step 4:

Then I would change the hydrogen on the right by putting a 4 in front of H₂O.

C5H7OH(l) + O2(g) —	—> 5 CO2(g) + 4 H2O(g)
C = 5	C = 5
H = 8	H = 8
O = 3	O = 14

 

Step 5:

Now we run into a problem. The oxygens are an even number on the right side and an odd number on the left side. To simplify my problem all we have to do is make the oxygens an even number on the left side (reactants). How do I do that? Change the coefficient of C₅H₇OH to 2.

2 C5H7OH(l) + O2(g) —	—> 5 CO2(g) + 4 H2O(g)
C = 10	C = 5
H = 16	H = 8
O = 4	O = 14

 

Step 6:

This unbalances my right side again so I go back to that side and fix it. Change coefficient in front of CO₂ to 10.

2 C5H7OH(l) + O2(g) —	—> 10 CO2(g) + 4 H2O(g)
C = 10	C = 10
H = 16	H = 8
O = 4	O = 24

 

Step 7:

Lets fix the hydrogen on the right by changing the coefficient of the H₂O to 8.

2 C5H7OH(l) + O2(g) —	—> 10 CO2(g) + 8 H2O(g)
C = 10	C = 10
H = 16	H = 16
O = 4	O = 28

 

Step 8:

Now we have to make the oxygen on the left side match the right. Remember that 2 oxygen are already contained in the first compound on the right side. So we can find out what the coefficient of O₂ by the oxygen on the right minus 2 divided by 2 = 28 – 2 / 2 = 13

2 C5H7OH(l) + 13 O2(g) —	—> 10 CO2(g) + 8 H2O(g)
C = 10	C = 10
H = 16	H = 16
O = 28	O = 28

 

Step 9:

All the elements are the same on both sides so I am done.

COMPLETE ANSWER: 2 C₅H₇OH_(l) + 13 O_2(g) ——> 10 CO_2(g) + 8 H₂O_(g)

 

PRACTICE PROBLEMS: Balance the chemical equations below.

 

Na3PO4(aq) + CaCl2(aq) ——> Ca3(PO4)2(s) + NaCl(aq)

Answer: 2 Na3PO4(aq) + 3 CaCl2(aq) ——> Ca3(PO4)2(s) + 6 NaCl(aq)

K2(SO3)(aq) + Mg3(PO4)2(s) ——> MgSO3(aq) + K3PO4(aq)

Answer: 3 K2(SO3)(aq) + Mg3(PO4)2(s) ——> 3 MgSO3(aq) + 2 K3PO4(aq)

C2H6(l) + O2(g) ——> CO2(g) + H2O(g)

Answer: 2 C2H6(l) + 7 O2(g) ——> 4 CO2(g) + 6 H2O(g)

C6H11OH(l) + O2(g) ——> CO2(g) + H2O(g)

Answer: 2 C6H11OH(l) + 17 O2(g) ——> 12 CO2(g) + 12 H2O(g)