Question

2 moles of propane reacts with 8 moles of oxygen in a combustion reaction. How many moles of carbon dioxide are formed? How much of the excess reactant is left over?

Answer 1

In the combustion reaction of propane with oxygen, 2 moles of propane react with 8 moles of oxygen to produce 6 moles of carbon dioxide. Oxygen is the limiting reactant, and none will be left over after the reaction.

Step-by-step explanation:

Stoichiometry of Propane Combustion

The combustion reaction for propane (C₃H₈) with oxygen (O₂) to form carbon dioxide (CO₂) and water (H₂O) is as follows:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

According to the balanced chemical equation, 1 mole of propane reacts with 5 moles of oxygen to produce 3 moles of carbon dioxide. Given that we have 2 moles of propane reacting with 8 moles of oxygen, we can calculate the number of moles of carbon dioxide formed as follows:

1. Moles of propane (C₃H₈) × (Moles of CO₂ produced per mole of C₃H₈) =
   2 mol × 3 mol CO₂/mol C₃H₈ = 6 mol CO₂

To determine the excess reactant and how much is left over, we must identify the limiting reactant:

1. Moles of oxygen required for complete combustion of 2 moles of propane =
   2 mol × 5 mol O₂/mol C₃H₈ = 10 mol O₂
   
2. 
   Since only 8 moles of O₂ were provided and 10 moles are required for complete combustion, oxygen is the limiting reactant.
   

Thus, no oxygen will be left over, and all 8 moles will be consumed in the reaction to produce 6 moles of CO₂.