Reaction Stoichiometry

A balanced chemical equation tells us how many reactant molecules are consumed and how many product molecules are created in a chemical reaction. However, they can also be interpreted as the number of moles of each substance involved in the reaction.

We also know that using atomic, molecular and formula weights we can convert between numbers of molecules or moles and masses of substances.

Combining these two concepts allows us to predict how much of each reactant is needed and how much of each product is produced. The ability to do this correctly is a fundamental aspect of experimental chemistry.

As an example of this let’s return to the combustion of octane in an automobile engine.

Example

What quantities (in grams) of carbon dioxide and water are produced from combustion of 2.8 kg of octane (~ 1 gallon)?

To begin the problem we need a balanced equation and the molecular/formula weights of the involved products and reactants.

2C8H18 + 25O2 ® 16CO2 + 18H2O

• MW C8H18 = 8(12.01) + 18(1.01) = 114.26 g/mol
• MW O2 = 2(16.00) = 32.00 g/mol
• MW CO2 = 12.01 + 2(16.00) = 44.01 g/mol
• MW H2O = 2(1.01) + 16.00 = 18.02 g/mol

From this point given the mass of one substance (A) we can calculate the mass of any other substance (B) in the reaction using the following strategy.

Grams A (MW A) ® Moles A (Equation) ® Moles B (MW B) ® Grams B

• To convert from grams of substance A to moles of substance A we use the molecular weight of substance A.
• To convert from moles of substance A to moles of substance B we use the balanced chemical equation.
• To convert from moles of substance B to grams of substance B we use the molecular weight of substance B.

2.8 kg C8H18 ´ (1000 g C8H18/1 kg C8H18) ´ (1 mol C8H18/114.26 g C8H18) ´ (16 mol CO2/2 mol C8H18) x (44.01 g CO2/1 mol CO2)

= 8.6´ 103 g CO2 = 8.6 kg CO2

2.8 kg C8H18 ´ (1000 g C8H18/1 kg C8H18) ´ (1 mol C8H18/114.26 g C8H18) ´ (18 mol H2O/2 mol C8H18) x (18.02 g H2O/1 mol H2O)

= 4.0´ 103 g H2O = 4.0 kg H2O

Next we might ask how much oxygen is needed to react with 2.8 kg of octane. Once again we repeat the same procedure.

2.8 kg C8H18 ´ (1000 g C8H18/1 kg C8H18) ´ (1 mol C8H18/114.26 g C8H18) ´ (25 mol O2/2 mol C8H18) x (32.00 g O2/1 mol O2)

= 9.8´ 103 g O2 = 9.8 kg O2

Limiting Reagent and Theoretical Yield

The concept of a limiting reagent is really quite a simple one. As the reaction is being carried out, each reactant is being consumed. The reaction will end as soon as one of the reactant runs out.

The reactant that runs out is called the limiting reagent.

The amount of product that is produced is called the theoretical yield.

Let’s illustrate the concept of a limiting reaction by considering an example far removed from chemistry.

Example

Lets say we wanted to start a series of 5 on 5 pick up full court basketball games at a local gym. There are 12 courts available 82 players and 21 basketballs.

Our equation for this reaction is

1 court + 10 players + 1 basketball ® 1 (5 ´ 5 Bball game)

There are various ways to carry out the problem. One way is to calculate how many games we could set up with each reactant, assuming we had an unlimited supply of the other reactants.

12 courts ´ (1 Bball Game/1 court) = 12 Bball games

82 players ´ (1 Bball Game/10 players) = 8.2 Bball games

21 balls ´ (1 Bball Game/1 ball) = 21 Bball games

So we will run out of players before balls or courts. Therefore, players are the limiting reagent and the theoretical yield is 8 Basketball games.

Now lets try a chemical example.

Example

Consider the following reaction

4NH3(g) + 5O2(g) ® 4NO(g) + 6H2O

If we react 1.50 g of ammonia with 1.85 g of oxygen, what is the limiting reagent and the theoretical yield of NO.

• MW NH3 = 17.04 g/mol
• MW O2 = 32.00 g/mol
• MW NO = 30.01 g/mol

We will answer the question in the same manner, by calculating how much NO would be produced from each reactant, provided we had an unlimited supply of the other reactant.

1.50 g NH3 ´ (1 mol NH3/17.04 g NH3) ´ (4 mol NO/4 mol NH3) ´ (30.01 g NO/1 mol NO)

= 2.64 g NO

1.85 g O2 ´ (1 mol O2/32.00 g O2) ´ (4 mol NO/5 mol NH3) ´ (30.01 g NO/1 mol NO)

= 1.39 g NO

The fact that 1.85 g of O2 would produce less NO than 1.50 g of NH3 means that the oxygen will run out first. Therefore, O2 is the limiting reagent and the theoretical yield is 1.39 g of NO.

Some other definitions that we can introduce here

Excess Reactant = All reactants which are not the limiting reactant are called excess reactants. They will not be fully consumed during the reaction.

Actual Yield = The actual amount of a product which is produced.

Percent Yield = (Actual Yield/Theoretical Yield) ´ 100

Example

If the actual yield of the above reaction was 1.11 g, what would the percent yield be?

Percent Yield = (1.11 g/1.39 g) ´ 100 = 79.9%

In class demo – Ignition of balloons filled with various ratios of H2 and O2