Stoichiometry
Mole-Mole Examples

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The solution procedure used below involves making two ratios and setting them equal to each other. When two ratios are set equal, this is called a proportion and the whole technique (creating two ratios, setting them equal) is called ratio-and-proportion.

One ratio will come from the coefficients of the balanced equation and the other will be constructed from the problem. The ratio set up from data in the problem will almost always be the one with an unknown in it.

Key point: the two ratios have to be set up with equivalent things in the same relative place in each ratio. A bit confusing? I will elaborate on this below.

After setting up the proportion, you will cross-multiply and divide to get the answer.

What happens if the equation isn't balanced? Answer: balance it. You cannot do these problems correctly without a balanced equation. The ChemTeam is constantly amazed at the number of people who forget to balance the equation first. One note: remember that there are chemical equations where all the coefficients are a value of one. These equations are already balanced. The term that is often used for these equations is "balanced as written."

How will I know which substances to use in the ratio? Answer: you will have to read the problem and understand the words in it.


Here is the first equation we'll use:

N2 + 3H2 ---> 2NH3

Example #1: If we have 2.00 mol of N2 reacting with sufficient H2, how many moles of NH3 will be produced?

Comments prior to solving the example

(a) The equation is already balanced.
(b) The ratio from the problem will have N2 and NH3 in it.
(c) How do you know which number goes on top or bottom in the ratios? Answer: it does not matter, except that you observe the next point ALL THE TIME.
(d) When making the two ratios, be 100% certain that numbers are in the same relative positions. For example, if the value associated with NH3 is in the numerator, then MAKE SURE it is in both numerators.
(e) Use the coefficients of the two substances to make the ratio from the equation.
(f) Why isn't H2 involved in the problem? Answer: the word "sufficient" removes it from consideration.

Solution:

1) We will use this ratio to set up the proportion:

NH3
––––
N2

2) That means the ratio from the equation is:

2
––
1

3) The ratio from the data in the problem will be:

x
––––
2.00

4) The proportion (setting the two ratios equal) is:

2   x
–– = ––––
1   2.00

5) Solving by cross-multiplying gives:

x = 4.00 mol of NH3 produced.

Comment: Notice how the ratio-and-proportion is written. Written in this manner:

x   2.00
–– = ––––
2   1

is equally correct. Just make sure to keep the two quantities associated with the NH3 and the two associated with the N2 on the same relative side. (The ChemTeam tends to not write the ratio and proportion in the style of the one just about, so you won't see it any more.)


Example #2: Suppose 6.00 mol of H2 reacted with sufficient nitrogen. How many moles of ammonia would be produced?

Solution:

1) Let's use this ratio to set up the proportion:

NH3
––––
H2

2) That means the ratio from the equation is:

2
––––
3

3) The ratio from the data in the problem will be:

x
––––
6.00

4) The proportion (setting the two ratios equal) is:

2   x
–– = ––––
3   6.00

5) Solving by cross-multiplying and dividing gives:

3x = 12.00 mol

x = 4.00 mol of NH3 produced


Example #3: We want to produce 2.75 mol of NH3. How many moles of nitrogen would be required?

Before the solution, a brief comment: notice that hydrogen IS NOT mentioned in this problem. If any substance ISN'T mentioned in the problem, then assume there is a sufficient quantity of it on hand. Since that substance isn't part of the problem, then it's not part of the solution.

Solution:

1) Let's use this ratio to set up the proportion:

NH3
––––
N2

2) That means the ratio from the equation is:

2
––
1

3) The ratio from the data in the problem will be:

2.75
––––
x

4) The proportion (setting the two ratios equal) is:

2.75   2
––– = ––
x   1

5) Solving by cross-multiplying and dividing (plus rounding off to three significant figures) gives:

x = 1.38 mol of N2 needed.

Here's the equation to use for the next three examples:

2H2 + O2 ---> 2H2O

Example #4: How many moles of H2O are produced when 5.00 moles of oxygen are used?

1) Here are the two substances in the molar ratio I used:

O2
––––
H2O

2) The molar ratio from the problem data is:

5.00
––––
x

3) The proportion to use is:

5.00   1
–––– = ––
x   2

x = 10.0 mol of H2O are produced


Example #5: If 3.00 moles of H2O are produced, how many moles of oxygen must be consumed?

1) Here are the two substances in the molar ratio I used:

O2
––––
H2O

2) The molar ratio from the problem data is:

x
––––
3.00

3) The proportion to use is:

x   1
–––– = ––
3.00   2

x = 1.50 mol of O2 was consumed


Example #6: How many moles of hydrogen gas must be used, given the data in example #5?

Solution #1:

1) Here are the two substances in the molar ratio I used:

H2
––––
O2

2) The molar ratio from the problem data is:

x
––––
1.50

3) The proportion to use is:

x   2
–––– = ––
1.50   1

x = 3.00 mol of H2 was consumed

Notice that the above solution used the answer from example #5. The solution below uses the information given in the original problem:

Solution #2: The H2 / H2O ratio of 2/2 could have been used also. In that case, the ratio from the problem would have been 3.00 over x, since you were now using the water data and not the oxygen data.


Example #7: Use the following equation:

C3H8 + 3O2 ---> 3CO2 + 4H2

(a) How many moles of O2 are required to combust 1.50 moles of C3H8?
(b) How many moles of CO2 are produced?
(c) How many moles of H2 are produced?

Solution to (a):

1) Use this ratio from the balanced chemical equation:

13

Note the style change!

2) Use this ratio from the problem:

1.50x

3) Set equal and solve:

13 = 1.50x

x = 4.50 mol

Solution to (b):

Since CO2 has the same coefficient as O2, the answer will be the same: 4.50 moles of CO2 will be produced.

Solution to (c):

14 = 1.50x

x = 6.00 mol


Example #8: CuSO4 5H2O (a hydrated compound) is strongly heated, causing the water to be released. How many moles of water are produced when 1.75 moles of CuSO4 5H2O is heated?

Solution:

1) The chemical equation of interest is this:

CuSO4 5H2O ---> CuSO4 + 5H2O

2) Every one mole of CuSO4 5H2O that is heated releases five moles of water. The ratio from the chemical equation is this:

15

3) The ratio from the problem data is this:

1.75x

4) Solving:

15 = 1.75x

x = 8.75 moles of water will be produced


Example #9: 2.50 moles of K2CO3 1.5H2O is decomposed. How many moles of water will be produced?

Solution:

23 = 2.50x

x = 3.75

Notice the use of a two-to-three ratio in place of a one-to-one-point-five ratio.


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