Consider the reaction below. if you start with 4.00 moles of c3h8 (propane) and 4.00 moles of o2 , how many moles of propane would be consumed?

0.800 mol

Explanation:

We have the amounts of two reactants, so this is a limiting reactant problem.  

We know that we will need a balanced equation with moles of the compounds involved.  

Step 1. Gather all the information in one place.

            C₃H₈ + 5O₂ ⟶ 3CO₂ + 4H₂O

n/mol:  4.00    4.00

Step 2. Identify the limiting reactant

Calculate the moles of CO₂ we can obtain from each reactant.  

From C₃H₈:

The molar ratio of CO₂: C₃H₈ is 3:1

Moles of CO₂ = 4.00 × 3/1

Moles of CO₂ = 12.0 mol CO₂

From O₂:

The molar ratio of CO₂: O₂ is 3:5.

Moles of CO₂ = 4.00 × ⅗

Moles of CO₂ = 2.40 mol CO₂

O₂ is the limiting reactant because it gives the smaller amount of CO₂.

Step 3. Calculate the moles of C₃H₈ consumed.

The molar ratio of C₃H₈:O₂ is 1:5.

Moles of C₃H₈ = 4.00 × ⅕

Moles of C₃H₈ = 0.800 mol C₃H₈