How do you solve for volume conversions and empirical formula?

Question

1. The combustion of propane gas produces carbon dioxide and water vapor.

                     C3H8(g)  +  5 O2(g)  →  3 CO2(g)  +  4 H2O(g)

(A) What volume of oxygen is required to completely combust 0.650 L of propane (C3H8) at STP?

(B) What volume of carbon dioxide is produced in the reaction?

Answer 1

a) At STP, 1 mole of any gas occupies 22.4 L: (0.650 L C₃H₈) * (1 mol C₃H₈ /22.4 L C₃H₈) * (5 mol O₂ /1 mol C₃H₈) * (22.4 L O₂ /1 mol O₂) = 3.25 L O₂.

b) For CO₂ the only difference is the mole ratio 3:1 CO₂ to C₃H₈, so the volume is 3*0.650 = 1.95 L CO₂.