Mole-Mole Examples

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:

N_{2}+ 3H_{2}---> 2NH_{3}

**Example #1:** If we have 2.00 mol of N_{2} reacting with sufficient H_{2}, how many moles of NH_{3} will be produced?

**Comments prior to solving the example**

(a) The equation is already balanced.

(b) The ratio from the problem will have N_{2}and NH_{3}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 NH_{3}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 H_{2}involved in the problem? Answer: the word "sufficient" removes it from consideration.

**Solution:**

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

NH _{3}–––– N _{2}

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 NH_{3}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 NH_{3} and the two associated with the N_{2} 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 H_{2} reacted with sufficient nitrogen. How many moles of ammonia would be produced?

**Solution:**

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

NH _{3}–––– H _{2}

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 molx = 4.00 mol of NH

_{3}produced

**Example #3:** We want to produce 2.75 mol of NH_{3}. 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:

NH _{3}–––– N _{2}

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 N_{2}needed.

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

2H_{2}+ O_{2}---> 2H_{2}O

**Example #4:** How many moles of H_{2}O are produced when 5.00 moles of oxygen are used?

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

O _{2}–––– H _{2}O

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 H

_{2}O are produced

**Example #5:** If 3.00 moles of H_{2}O are produced, how many moles of oxygen must be consumed?

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

O _{2}–––– H _{2}O

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 O

_{2}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:

H _{2}–––– O _{2}

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 H

_{2}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 H_{2} / H_{2}O 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:

C_{3}H_{8}+ 3O_{2}---> 3CO_{2}+ 4H_{2}

(a) How many moles of O_{2} are required to combust 1.50 moles of C_{3}H_{8}?

(b) How many moles of CO_{2} are produced?

(c) How many moles of H_{2} are produced?

**Solution to (a):**

1) Use this ratio from the balanced chemical equation:

^{1}⁄_{3}Note the style change!

2) Use this ratio from the problem:

^{1.50}⁄_{x}

3) Set equal and solve:

^{1}⁄_{3}=^{1.50}⁄_{x}x = 4.50 mol

**Solution to (b):**

Since CO_{2}has the same coefficient as O_{2}, the answer will be the same: 4.50 moles of CO_{2}will be produced.

**Solution to (c):**

^{1}⁄_{4}=^{1.50}⁄_{x}x = 6.00 mol

**Example #8:** CuSO_{4} **⋅** 5H_{2}O (a hydrated compound) is strongly heated, causing the water to be released. How many moles of water are produced when 1.75 moles of CuSO_{4} **⋅** 5H_{2}O is heated?

**Solution:**

1) The chemical equation of interest is this:

CuSO_{4}⋅5H_{2}O ---> CuSO_{4}+ 5H_{2}O

2) Every one mole of CuSO_{4} **⋅** 5H_{2}O that is heated releases five moles of water. The ratio from the chemical equation is this:

^{1}⁄_{5}

3) The ratio from the problem data is this:

^{1.75}⁄_{x}

4) Solving:

^{1}⁄_{5}=^{1.75}⁄_{x}x = 8.75 moles of water will be produced

**Example #9:** 2.50 moles of K_{2}CO_{3} **⋅** 1.5H_{2}O is decomposed. How many moles of water will be produced?

**Solution:**

^{2}⁄_{3}=^{2.50}⁄_{x}x = 3.75

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