### StoichiometryMole-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:

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! 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:

1.50x

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.

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

5C(s) + 2SO2(g) ---> CS2(ℓ) + 4CO(g)

(a) How many moles of carbon are needed to react with 5.01 mol SO2?
(b) How many moles of carbon monoxide form at the same time that 0.255 mol SO2 forms?
(c) How many moles of SO2 are required to make 125 mol CS2?
(d) How many moles of CS2 form when 4.1 mol C reacts?
(e) How many moles of carbon monoxide form at the same time that 0.255 mol CS2 forms?

Solution to (a):

The molar ratio between C and SO2 is 5:2.

The ratio and proportion to be used is this:

5 is to 2 as x is to 5.01

x = 12.5 mol (to three sig figs)

Solution to (b):

The molar ratio between CO and SO2 is 4:2.

The ratio and proportion to be used is this:

4 is to 2 as x is to 0.255

x = 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 SO2:CS2 molar ratio is 2:1

The 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:CS2 is 5:1

5 is to 1 as 4.1 is to x

x = 0.82 mol (to two sig figs)

Solution to (e):

CO:CS2 molar ratio is 4:1

 4 x –– = –––– 1 0.255

x = 1.02 mol

Example #11: 2NO(g) + O2(g) ---> 2NO2(g)

(a) How many moles of O2 combine with 500. moles of NO?
(b) How many moles of NO2 are formed from 0.250 mole of NO and sufficient O2?
(c) How many moles of O2 are left over if 80.0 moles of NO is mixed with 200. moles of O2 and the mixture reacts?

Solution to (a):

NO and O2 react in a 2:1 molar ratio

 2 500. mol –––– = ––––––– 1 x

x = 250. mol

Solution to (b):

NO and NO2 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 O2 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 O2 left over.

Example #12: Consider the reaction:

4Al(s) + 3O2(g) ---> 2Al2O3(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 Al2O3 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 Al2O3 molar ratio is 2:1

 2 x –––– = ––––––––– 1 0.438 mol

x = 0.876 mol of Al required.

Second, oxygen

O2 to Al2O3 molar ratio is 3:2

 3 x –––– = ––––––––– 2 0.438 mol

x = 0.657 mol of oxygen required

Solution to (c):

Al to O2 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 #13: 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:

2H3PO4(ℓ) + 3CaCl2(aq) ---> Ca3(PO4)2(s) + 6HCl(aq)

2) Part (a)

CaCl2 to Ca3(PO4)2 molar ratio is 3:1

 3 x –––– = ––––––––– 1 0.570 mol

x = 1.71 mol of CaCl2 required.

2) Part (b)

H3PO4 to CaCl2 molar ratio is 2:3

 2 x –––– = ––––––––– 3 1.37 mol

x = 0.667 mol of H3PO4 required.