What is stoichiometry? It is the relationship between the amounts of reactants and products in a balanced chemical equation. The coefficients in the chemical reaction are called stoichiometric coefficients. For the following balanced chemical equation, 1 formula unit of gaseous iodine reacts with 1 formula unit of hydrogen gas to produce 2 formula units of gaseous hydrogen iodide. It can also be expressed in terms of moles: 1 mole of iodine gas reacts with 1 mole gaseous hydrogen to produce 2 moles of hydrogen iodide gas.

I₂ (g) + H₂ (g) → 2 HI (g)

In the example below, we have 1 mole of nitrogen gas reacting with 3 moles of hydrogen gas to produce 2 moles of gaseous ammonia.

N₂ (g) + 3 H₂ (g) → 2 NH₃ (g)

How many moles of NH₃ can be produced if 2 moles of N₂ react? From the balanced chemical equation, there is a 1:2 ratio of N₂:NH₃ meaning 4 moles of NH₃ would be produced. How many moles of NH₃ can be produced if 6 moles of H₂ reacts? The answer is 4 moles of NH₃. There is a 3:2 ratio of H₂:NH₃.

Using the balanced equation we can set up mole ratios to help us with stoichiometry calculations. How many moles of NH₃ can be formed if 25.5 moles of N₂ react? The mole ratios of N₂ and NH₃ follow:

\(\displaystyle\frac{1\;mol\;N_2}{2\;mol\;NH_3}\qquad or \qquad\frac{2\;mol\;NH_3}{1\;mol\;N_2}\)

We multiply the 25.5 moles N₂ by the mole ratio which will cancel out the number of moles of N₂–in this case, the second mole ratio.

\(\displaystyle \require{cancel}25.5\;\cancel{mol\;N_2}\times\frac{2\;mol\;NH_3}{1\cancel{\;mol\;N_2}}\;=\;\mathbf{51.0\;mol\;NH_3}\)

In the lab, we work with mass. Using the same chemical equation,

N₂ (g) + 3 H₂ (g) → 2 NH₃ (g)

calculate the number of grams of hydrogen that are required to react with 10.5 g of N₂.

First, write the mole ratios that relate the number of moles of hydrogen and nitrogen.

\(\displaystyle\frac{1\;mol\;N_2}{3\;mol\;H_2}\qquad or \qquad\frac{3\;mol\;H_2}{1\;mol\;N_2}\)

Next, determine the molar mass of N₂ and H₂. The molar mass of N₂ is 28.034 g/mol and for H₂ is 2.0159 g/mol. The grams of N₂ must be converted to moles. We then use the mole ratio to cancel out N₂ and leave us with moles of H₂. Finally, convert the moles of H₂ to grams. Here is the roadmap:

g N₂ → mol N₂ → mol H₂ → g H₂

We have three arrows and one equivalence per arrow.

\(\displaystyle10.5\;{g\;N_2}\;\times\;\frac{1\;mol\;N_2}{28.034\;g\;N_2}\;\times\;\frac{3\;mol\;H_2}{1\;mol\;N_2}\;\times\;\frac{2.0159\;g\;H_2}{1\;mol\;H_2}\;=\;\mathbf{2.27\;g\;H_2}\)

In the problems we just completed, we were given an amount of reactant or an amount of produce. For most chemical reactions, there is a Limiting Reactant (or limiting reagent), which is the amount of reactant that limits the amount of product formed. Suppose we run an auto industry where we are responsible for tires and steering wheels to produce a very basic car.

We see that for each car, we need one steering wheel and four tires. How many cars can be produced if we have 300 steering wheels and 12 tires? Obviously, we can only produce 3 cars because the number of tires limit the number of cars we can manufacture. You can think of the tires as the limiting reactant and the steering wheels as being in excess or our excess reactant. In fact, once the three cars are produced, 297 steering wheels will remain unused. If we have 75 steering wheels and 500 tires, we can produce 75 cars. In this case, the steering wheels are limiting and and the tires are in excess.

We can apply the same principles to chemical reactions. Consider the following chemical equation for the production of hydroiodic acid. Calculate the number of grams of HI that can be produced when 65.5 g of I₂ is reacted with 55.5 g of N₂H₄.

I₂ + N₂H₄ → HI + N₂

First, the equation must be balanced.

2 I₂ + N₂H₄ → 4 HI + N₂

The easiest way to do the problem is to determine the number of moles of hydroiodic acid that is produced when each reactant completely reacts. Once that is done, we can determine the limiting reactant. First we start with 65.5 g of I₂. Write the mole ratios for iodine and hydroiodic acid.

\(\displaystyle\frac{2\;mol\;I_2}{4\;mol\;HI}\qquad or \qquad\frac{4\;mol\;HI}{2\;mol\;I_2}\)

The molar mass of I₂ is 253.808 g/mol.

\(\displaystyle 65.5\;{g;I_2}\times\frac{1\;mol\;I_2}{253.808\;g}\times\frac{4\;mol\;HI}{2\;mol\;I_2}\;=\;\mathbf{0.516\;mol\;HI}\)
 
If 65.5 g of I₂ reacts, 0.516 moles of HI would be produced. Next we calculate the moles of HI produced if 55.5 g of N₂H₄ completely reacts. The molar mass of N₂H₄ is 32.0452 g/mol. The mole ratios of N₂H₄ and HI are:

\(\displaystyle\frac{1\;mol\;H_2N_4}{4\;mol\;HI}\qquad or \qquad\frac{4\;mol\;HI}{1\;mol\;H_2N_4}\)

Now we determine the number of moles of HI that are produced when 55.5 g of N₂H₄ react.

\(\displaystyle 55.5\;g\;N_2H_4\;\times\frac{1\;mol\;N_2H_4}{32.0452\;g\;N_2H_4}\times\frac{4\;mol\;HI}{1\;mol\;N_2H_4}\;=\;\mathbf{6.93\;mol\;HI}\)

We can see that I₂ is the limiting reactant. When 65.5 g of I₂ is reacted with 55.5 g of N₂H₄ the most HI that can be produced is 0.516 moles. Finally, we convert 0.516 moles of HI to grams. The molar mass of HI is 127.91 g/mol.

\(\displaystyle 0.516\;mol\;HI\;\times\;\frac{127.91\;g\;HI}{1\;mol\;HI}\;=\;\mathbf{66.0\;g\;HI}\)

The amount of HI that can be produced is 66.0 g HI. This is known as the theoretical yield. The theoretical yield is the calculated yield of product. In the lab, when running an experiment, there are experimental errors. The percent yield is calculated from the actual yield and the theoretical yield. The actual yield is the amount of product formed when the experiment is run in the lab. For the above example, let’s say we recovered 62.3 g of HI in the lab–the actual yield. The theoretical yield is 66.0 g HI. The percent yield would be calculated as:

\(\displaystyle Percent\;{yield}\;=\;\frac{actual\;yield}{theoretical\;yield}\;\times\;{100}\)

\(\displaystyle\frac{62.3\;g}{66.0\;g}\;\times\;{100}\;=\;\mathbf{94.4\;\%}\)

The percent yield is under 100% which is expected. Experimental errors and side reactions can affect the percent yield. Most chemists try to improve the percent yield of a reaction.

Continuing with the problem, we want to determine how much I₂ is left unreacted. There are several ways to do this by using mole ratios. The balanced equation is:

2 I₂ + N₂H₄ → 4 HI + N₂

We started out with 65.5 g of I₂ and 55.5 g of N₂H₄. We can calculate the number of moles of N₂H₄ required to produce 0.516 moles of HI by using mole ratios from the balanced chemical equation. We then convert the number of moles of N₂H₄ to grams and subtract that amount from the initial number of grams of N₂H₄, 55.5 g.

\(\displaystyle 0.516\;mol\;HI\;\times\frac{1\;mol\;N_2H_4}{4\;mol\;HI}\times\frac{32.0452\;g\;N_2H_4}{1\;mol\;N_2H_4}\;=\;\mathbf{4.13\;g\;N_2H_4\;reacted}\)

The amount of N₂H₄ left unreacted is:

55.5 g – 4.13 g = 51.4 g N₂H₄

Worksheet: Stoichiometry Part 1
Worksheet: Stoichiometry Part 2
Worksheet: Stoichiometry Part 3
Worksheet: Atomic Mass and Molar Mass Conversions and Stoichiometry

Please watch the following videos before attempting the exercises

Stoichiometry Part 1

Stoichiometry Part 2

Stoichiometry Part 3

Stoichiometry Part 4

 

Exercises

View Solutions/Answers to Exercises

Exercise 1. Consider the following unbalanced equation:

Fe₂O₃ (s) + C → Fe (s) + CO₂ (g)

 

View Solutions/Answers to Exercise 1.

Exercise 2. Consider the following unbalanced chemical equation. How many grams of hydrogen are required to react with 243 g of nitrogen?

N₂ (g) + H₂ (g) → NH₃ (g)

View Solutions/Answers to Exercise 2.

Exercise 3. Solid nickel reacts with aqueous hydrochloric acid to produce aqueous nickel(II) chloride and hydrogen gas.

 

View Solutions/Answers to Exercise 3.

Exercise 4. Oxygen gas can be prepared by heating mercury(II) oxide according to the following equation:

\(\displaystyle HgO\;(s)\;\rightarrow\;\;Hg\;(l)\;+\;O_2\;(g)\)

How many grams of oxygen was formed if 535 g of HgO was heated?

View Solutions/Answers to Exercise 4.

Exercise 5.

Consider the reaction of aluminum hydroxide with sulfuric acid.

Al(OH)₃ (s) + H₂SO₄ (aq) → Al₂(SO₄)₃ (aq) + H₂O (l) (unbalanced)

If 36.52 g of aluminum hydroxide and 40.65 g of sulfuric acid are allowed to react, how many grams of aluminum sulfate will form?

View Solutions/Answers to Exercise 5.

Exercise 6.

Benzene, C₆H₆ reacts with bromine, in the presence of iron, to give bromobenzene, C₆H₅Br.

C₆H₆ (l) + Br₂ (l) → C₆H₅Br (l) + HBr (g)

What is the percent yield if 126 g of C₆H₅Br is recovered when 165 g of C₆H₆ is reacted with 165 g of Br₂?

View Solutions/Answers to Exercise 6.

Exercise 7.

Gaseous hydrogen sulfide, H₂S is bubbled into a solution of sodium hydroxide to produce aqueous sodium sulfide and liquid water. Assuming the percent yield of sodium sulfide is 94.0%, how many grams of sodium sulfide are formed if 2.5 g of H₂S is bubbled into a solution that contains 4.00 g of NaOH? Make sure you write a balanced chemical equation.

View Solutions/Answers to Exercise 7.

 

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