What are the types of stoichiometry examples, with examples?

1 Answer
Jan 15, 2014

Mole-mole; mass-mole; volume-mole; mass-mass; mass-volume; volume-volume

Stoichiometry problems are usually classified according to the measurements used for the reactants involved — moles, mass, and volume.

Here are some examples of the types of problems you will encounter. Mole-mole conversions are at the heart of every stoichiometry calculation.


Sulfur reacts with oxygen to form sulfur trioxide according to the equation

2S + 3O₂ → 2SO₃

How many moles of sulfur react with 9.00 mol O₂?

9.00 mol O₂ × #(2 mol S)/(3 mol O₂)# = 6.00 mol S


Example 1

Oxygen is produced by the decomposition of potassium chlorate according to the equation

2KClO₃ →2KCl + 3O₂

How many moles of oxygen are produced by the decomposition of 15.0 g of potassium chlorate?

Here, you must convert grams of KClO₃ to moles of KClO₃ before you can do a mole-mole conversion.

15.0 g KClO₃ × #( 1 mol KClO₃)/(122.6 g KClO₃) × (3 mol O₂)/(2 mol KClO₃)# = 0.184 mol O₂

Example 2

What mass of KClO₃ must be decomposed in order to produce 0.200 mol O₂?

Here, we do the mole-mole conversion first and then do a mole-mass conversion.

0.200 mol O₂ × #(2 mol KClO₃)/(3 mol O₂) × (122.6 g KClO₃)/(1 mol KClO₃)# = 16.3 g KClO₃


Example 1

Hydrogen and nitrogen react to form ammonia according to the equation

N₂ + 3H₂ → 2NH₃

How many moles of hydrogen are needed to produce 224 L NH₃?

Here, we must do a volume-mole conversion before the mole-mole conversion. The conversion factor at STP is

#(22.414L)/(1 mol)# or #(1 mol)/(22.414 L)#

224 L NH₃ × #(1 mol NH₃)/(22.414 L NH₃) × (3 mol H₂)/(2 mol NH₃)# = 15.0 mol H₂

If you are given the volume at some other temperature and pressure, you will have to use the Ideal Gas Law to calculate the number of moles.

Example 2

What volume of NH₃ is formed from 15.0 mol H₂?

15.0 mol H₂ × #(2 mol NH₃)/(3 mol H₂) × (22.414 L NH₃)/(1 mol NH₃)# = 224 L NH₃


What mass of chlorine can be formed by the decomposition of 64.0 g of AuCl₃ by the following reaction?

2AuCl₃ → 2 Au + 3Cl₂

64.0 g AuCl₃ × #(1 mol AuCl₃)/(303.3 g AuCl₃) × (3 mol Cl₂)/(2 mol AuCl₃) × (70.91 g Cl₂)/(1 mol Cl₂)# = 22.4 g Cl₂


Example 1

What volume of carbon dioxide at 1.00 atm and 112.0 °C will be produced when 80.0 g of methane is burned?

CH₄ + 2O₂ → CO₂ + 2H₂O

80.0 g CH₄ × #(1 mol CH₄)/(16.04 g CH₄) × (1 mol CO₂)/(1 mol CH₄)# = 4.99 mol CO₂

We now use the Ideal Gas Law to calculate the volume of CO₂.

PV = nRT

V = #(nRT)/P#

n = 4.99 mol; R = 0.082 06 L•atm•K⁻¹mol⁻¹; 1.00 atm;
T = (112.0 + 273.15) K = 385.2 K; P = 1.00 atm

V =

#(4.99 mol × 0.082 06 L•atm•K⁻¹mol⁻¹ × 385.2 K × 1.00 atm)/(1.00 atm)# =

158 L

Example 2

What mass of CO₂ is formed by the combustion of 160 L CH₄ at 1.00 atm and 112.0 °C?

We must first use the Ideal Gas Law to calculate the moles of CH₄.

PV = nRT

n = #(PV)/(RT)#

n = #(1.00 atm × 160 L)/(0.08206L•atm•K⁻¹mol⁻¹ × 385.2 K)# = 5.06 mol CH₄

5.06 mol CH₄ × #(1 mol CO₂)/(1 mol CH₄) × (44.01 g CO₂)/(1 mol CO₂)# = 223 g CH₄


N₂ + 3H₂ → 2NH₃

What volume of hydrogen is necessary to react with 5.00 L of nitrogen to produce ammonia?

We would normally use the conversions

V of N₂ → moles of N₂ → moles of H₂→ V of H₂

The problem doesn’t give us the temperature or the pressure of the gases. However, we can use a trick. We know that 1 mol of any gas has the same volume as 1 mol of any other gas at the same temperature and pressure. Therefore, the volume ratios are the same as the molar ratios. We can write

5.00 L N₂× #(3 L H₂)/(1 L N₂)# = 15.0 L H₂

Of course, if the gases had been at different temperatures, then we would have had to use the Ideal Gas Law to get the volume-mole conversions.