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? Then your first step is to 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 have a value of one. These equations are already balanced. The term that is often used for these equations is "balanced as written."

Here is the first equation we'll use:

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

**Example #1:** When 2.00 mol of N_{2} reacts 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 mol NH _{3}–––––––– 1 mol N _{2}

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

x –––––––––– 2.00 mol N _{2}

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

2 mol NH _{3}x <--- both values in the numerator are related to ammonia ––––––––– = –––––––––– 1 mol N _{2}2.00 mol N _{2}<--- both values in the denominator are related to nitrogen

5) Solving by cross-multiplying and dividing gives:

(1 mol N_{2}) (x) = (2 mol NH_{3}) (2 mol N_{2})

(2 mol NH _{3}) (2.00 mol N_{2})x = ––––––––––––––––––––– 1 mol N _{2}x = 4.00 mol NH

_{3}produced.

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

x 2.00 mol N _{2}–––––––– = ––––––––– 2 mol NH _{3}1 mol N _{2}

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 side.

The ChemTeam tends to not write the ratio and proportion in the style of the one just above, 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 mol NH _{3}–––––––– 3 mol H _{2}

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

x –––––––––– 6.00 mol H _{2}

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

2 mol NH _{3}x –––––––– = –––––––––– <--- the two ammonia values are in the numerators and the two hydrogen value are in the denominator 3 mol H _{2}6.00 mol H _{2}

5) Solving by cross-multiplying and dividing gives:

3x = 12.00 molx = 4.00 mol 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 mol NH _{3}––––––––– 1 mol N _{2}

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

2.75 mol NH _{3}–––––––––– x

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

2.75 mol NH _{3}2 NH _{3}––––––––––– = ––––––– x 1 mol N _{2}

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

x = 1.38 mol 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}1 mol O _{2}–––– and the ratio is ––––––––– H _{2}O2 mol H _{2}O

2) The molar ratio from the problem data is:

5.00 mol O _{2}––––––––– x

3) The proportion to use is:

5.00 mol O _{2}1 mol O _{2}–––––––––– = ––––––––– x 2 mol H _{2}Ox = 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}OIt's a 1:2 ratio.

2) The molar ratio from the problem data is:

x ––––––––––– 3.00 mol H _{2}O

3) The proportion to use is:

x 1 mol O _{2}–––––––––––– = –––––––– 3.00 mol H _{2}O2 mol H _{2}Ox = 1.50 mol of O

_{2}consumed

For the examples below, I left off the mol unit on the ratio from the coefficients of the balanced equation. Also, I used a different way to format the ratios and the proportional set up.

**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! Just by the by, students often are confused when they see information presented to them in a different (but mathematically equivalent) style. Be aware!

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.

**Example #10:** Carbon disulfide is an important industrial solvent. It is prepared by the reaction of carbon (called coke) with sulfur dioxide:

5C(s) + 2SO_{2}(g) ---> CS_{2}(ℓ) + 4CO(g)

(a) How many moles of carbon are needed to react with 5.01 mol SO_{2}?

(b) How many moles of carbon monoxide form at the same time that 0.255 mol SO_{2} forms?

(c) How many moles of SO_{2} are required to make 125 mol CS_{2}?

(d) How many moles of CS_{2} form when 4.1 mol C reacts?

(e) How many moles of carbon monoxide form at the same time that 0.255 mol CS_{2} forms?

**Solution to (a):**

The molar ratio between C and SO_{2}is 5:2.The ratio and proportion to be used is this:

5 is to 2 as x is to 5.01x = 12.5 mol (to three sig figs)

**Solution to (b):**

The molar ratio between CO and SO_{2}is 4:2.The ratio and proportion to be used is this:

4 is to 2 as x is to 0.255x = 0.510 mol (to three sig figs)

**Short commentary:** when I solved part b, I simply multiplied 0.255 by 2. You may ponder why that was so. Also, when I solved all the problems in this example, I went to a piece of paper and wrote the ratio and proportion thusly [using (b) for an example]:

4 x –– = –––– 2 0.255

For myself personally, it gives me a better feel for solving the problem to look at the above formulation as opposed to this:

4 is to 2 as x is to 0.255

You need to be able to translate the "in a line" style to the "ratios written as fractions" style.

Lastly, note how in (b), I used the compounds "right to left" from the chemical equation as opposed to the other three where the reading of the compounds is "left to right." It's just a stylistic thing, but I do tend more to the "left to right" reading.

**Solution to (c):**

The SO_{2}:CS_{2}molar ratio is 2:1The proper ratio and proportion is this:

2 x –– = –––– 1 125 x = 250. (to three sig figs, note the explicit decimal point)

**Solution to (d):**

C:CS_{2}is 5:15 is to 1 as 4.1 is to x

x = 0.82 mol (to two sig figs)

**Solution to (e):**

CO:CS_{2}molar ratio is 4:1

4 x –– = –––– 1 0.255 x = 1.02 mol

**Example #11:** 2NO(g) + O_{2}(g) ---> 2NO_{2}(g)

(a) How many moles of O_{2} combine with 500. moles of NO?

(b) How many moles of NO_{2} are formed from 0.250 mole of NO and sufficient O_{2}?

(c) How many moles of O_{2} are left over if 80.0 moles of NO is mixed with 200. moles of O_{2} and the mixture reacts?

**Solution to (a):**

NO and O_{2}react in a 2:1 molar ratio

2 500. mol –––– = ––––––– 1 x x = 250. mol

**Solution to (b):**

NO and NO_{2}are in a 2:2 molar ratio

2 0.250 mol –––– = –––––––– 2 x x = 0.250 mol

Comment: be aware that a 2:2 ratio is the same as a 1:1 ratio. Often, a teacher will use a 1:1 ratio and students will become confused. "Where did the one-to-one ratio come from?" The answer is that a two-to-two ratio reduces to a one-to-one ratio. The teacher simply reduced it without mentioning it.

**Solution to (c):**

We first need to determine how many moles of oxygen are used when 80.0 moles of NO reacts.NO and O

_{2}react in a 2:1 molar ratio.

2 80.0 mol –––– = –––––––– 1 x x = 40.0 mol

Now, we can determine how much oxygen remains after the NO is used up.

200. − 40.0 = 160. mol of O

_{2}left over.

**Example #12:** Consider the following reaction:

4Al(s) + 3O_{2}(g) ---> 2Al_{2}O_{3}(s)

(a) Write the 6 mole ratios that can be derived from this equation. Write the first using the chemical formulas and, secondly, using the coefficients of the equation.

(b) How many moles of aluminum are needed to form 3.75 mol Al_{2}O_{3}?

**Solution to (a):**

Al:O _{2}Al to O_{2}Al/O_{2}Al:Al _{2}O_{3}

**Example #13:** Consider the reaction:

4Al(s) + 3O_{2}(g) ---> 2Al_{2}O_{3}(s)

(a) If 8.00 moles of aluminum react with an excess of oxygen, how many moles of aluminum oxide are produced?

(b) The production of 0.438 moles of aluminum oxide requires the reaction of ______ moles of aluminum and _____ moles of oxygen?

(c) When 1.830 moles of aluminum reacts, ______ moles of oxygen are consumed.

**Solution to (a):**

The molar ratio between Al and Al_{2}O_{3}is 2:1.Note that I reduced it from 4:2

2 8.00 mol –––– = ––––––– 1 x x = 4.00 mol

**Solution to (b):**

First, aluminum

Al to Al_{2}O_{3}molar ratio is 2:1

2 x –––– = ––––––––– 1 0.438 mol x = 0.876 mol of Al required.

Second, oxygen

O_{2}to Al_{2}O_{3}molar ratio is 3:2

3 x –––– = ––––––––– 2 0.438 mol x = 0.657 mol of oxygen required

**Solution to (c):**

Al to O_{2}molar ratio is 4:3

4 1.830 mol –––– = ––––––––– 3 x x = 1.3725 mol

To, four sig figs, this is 1.372 mol (the rule for rounding with a five applies).

**Example #14:** In a chemical reaction between phosphoric acid and aqueous calcium chloride, the products are hydrochloric acid and a precipitate of calcium phosphate.

(a) How many moles of calcium chloride are required to react in order to produce 0.570 moles of calcium phosphate?

(b) How many moles of phosphoric acid are required to react with 1.37 moles of calcum chloride?

**Solution:**

1) Write a balanced chemical equation:

2H_{3}PO_{4}(ℓ) + 3CaCl_{2}(aq) ---> Ca_{3}(PO_{4})_{2}(s) + 6HCl(aq)

2) Part (a)

CaCl_{2}to Ca_{3}(PO_{4})_{2}molar ratio is 3:1

3 x –––– = ––––––––– 1 0.570 mol x = 1.71 mol of CaCl

_{2}required.

2) Part (b)

H_{3}PO_{4}to CaCl_{2}molar ratio is 2:3

2 x –––– = ––––––––– 3 1.37 mol x = 0.913 mol of H

_{3}PO_{4}required.

**Example #15:** Given the reaction: 4NH_{3}(g) + 5O_{2}(g) ---> 4NO(g) + 6H_{2}O(ℓ)

When 1.20 mole of ammonia reacts, the total number of moles of products formed is:

(a) 1.20

(b) 1.50

(c) 1.80

(d) 3.00

(e) 12.0

**Solution:**

The correct answer is d.The NH

_{3}/ (NO + H_{2}O) molar ratio is 4:104 / 10 = 1.20 / x

x = 3.00 mol