I need help with my stoichiometry chemistry homework

Question

C3Hg +502 - 3C02 + 4H20

Carbon dioxide will be released from the burning of propane. Given 4.59 moles of propane and 3.85 moles of oxygen, use the equation to determine the limiting reactant, excess reactant, theoretical yield, and excess reactant leftover for the attached questions.

Answer 1

Propane is C₃H₈ (as opposed to C₃Hg that you listed)

Balanced chemical reaction equation:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

This tells you that for each mole of propane, you need 5 moles of oxygen. Can you see that since you have 4.59 moles of propane and only 3.85 moles of oxygen that you have way too much propane for the amount of oxygen you have? That means oxygen is your limiting reactant and that propane is your excess reactant (you have extra propane that will not burn)

So, when doing the stoichiometry, you only need to consider the amount of oxygen.

The balanced chemical reaction equation then tells you that for the 3.85 moles of oxygen, you will burn up 3.85/5 or 0.77 moles of propane, and you will get as a product 3.85(3/5) or 2.31 moles of CO₂ and 3.85(4/5) or 3.08 moles of water as your theoretical yield.

Since you are going to burn up 0.77 moles of propane and you start with 4.59 moles, you will have 4.59 - 0.77 or 3.82 moles of excess propane after the reaction.