Limiting Reactant

Question

In a reactions of propane (C3H8) and oxygen O2, and water and carbon dioxide were the products, which one of the following would be the limiting reactant if we used 10 grams of propane and 30 grams of oxygen?

Answer 1

First write the correctly balanced equation for the combustion of propane:

C₃H₈ + 5O₂ ==> 3CO₂ + 4H₂O

Next, find out how many moles of each reactant are present:

moles C₃H₈ = 10 g x 1 mole/44 g = 0.227 moles C₃H₈

moles O₂ = 30 g x 1 mole/32 g = 0.938 moles O₂

Just because C₃H₈ has fewer moles does NOT mean it is limiting, because we need to look at the mole ratios in the balanced equation:

0.227 moles C₃H₈ divided by its coefficient of 1 = 0.227 moles

0.938 moles O₂ divided by its coefficient of 5 = 0.188 moles

So, O₂ is limiting