To develop the principles for dealing with the stoichiometry of reactions, we will consider the reaction of propane with oxygen to produce carbon dioxide and water. We will consider the question: “What mass of oxygen will react with 96.1 g of propane?” In doing stoichiometry, the first thing we must do is write the balanced chemical equation for the reaction. In this case the balanced equation is

The above balanced equation mean that 1 mole of C₃H₈ will react with 5 moles of O₂ to produce 3 moles CO₂ and 4 moles H₂O. If the 1 molar of propane is 44.1 g/mol then the number of moles of propane:

1 mol propane needs 5 mol oxygen then the mole ratio is:

By multiplying with number of mole of propane we get number of mole of oxygen:

By multiplying with molar mass of O₂ we get the mass:

Now, what the mass of CO₂ produced from 96.1 g of propane?

By multiplying suitable mole ratio with number of mole of propane we get number of mole of CO₂:

By multiplying with molar mass of CO₂ we get the mass:

**Problem-Solving Strategy**

Calculating Masses of Reactants and Products in Chemical Reactions

1. Balance the equation for the reaction.

2. Convert the known mass of the reactant or product to moles of that substance.

3. Use the balanced equation to set up the appropriate mole ratios.

4. Use the appropriate mole ratios to calculate the number of moles of the desired reactant or product.

5. Convert from moles back to grams if required by the problem.