How do you solve for volume to 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

Hello, Jaida,

Thanks for the balanced equation. Stanton is, again, correct. Let's start by converting 0.650 liters of propane to moles propane. As described in the earlier problem, we can use the conversion factor 22.4 liters/mole of any gas at STP.

(0.650 L propane)/(22.4 L/mole) = 0.0290 moles propane.

We need 5moles of oxygen for every 1 mole propane, so:

5*(0.0290 moles propane) = 0.145 moles O₂

To find volume O₂ (at STP) multiply by 22.4 Liters/mole

We need 3.25 liters of oxygen at STP to completely react with all the propane.

One mole of propane produces 3 moles of carbon dioxide. So multiply the moles propane by 3 to find moles CO₂. That gives us 0.087 moles of carbon dioxide. Convert to volume CO₂ by multiplying by 22.4 liters/mole. (All STP)

3 sig figs for the final answer.

Bob