Limiting Stoichiometry with showing steps

Question

4.35 L of Butane (C4H10) is combusted with 26.50 L of Oxygen gas. What volume of each product is formed by this reaction? What volume of excess reactant remains?

Answer 1

First, always write the correctly balanced equation for the reaction taking place:

2C₄H₁₀ + 13O₂ ==> 8CO₂ + 10H₂O ... balanced equation

Next, find the limiting reactant. An easy way to do this is to simply divide the mols of each reactant by the corresponding coefficient in the balanced equation. In the current problem we are given liters of each reactant which for all intents and purpose is the same as moles as they become interchangeable when speaking about gases.

C₄H₁₀: 4.35 L ÷ 2 = 2.18

O₂: 26.50 L ÷ 13 = 2.04

Since 2.04 is less than 2.18, O₂ is the limiting reactant and will determine the maximum amount of each product that can be formed

Next, we use the stoichiometry of the balanced equation along with dimensional analysis to find the volume of each product (gas) that will be formed.

26.50 L O₂ x 8 L CO₂ / 13 L O₂ = 16.3 L CO₂ formed

26.50 L O₂ x 10 L H₂O / 13 L O₂ = 20.4 L H₂O formed

Finally, to find the volume of excess reactant (C₄H₁₀) that remains, we again use the stoichiometry of the balanced equation to find volume of C₄H₁₀ used up, and then subtract that from the volume we started with.

26.50 L O₂ x 2 L C₄H₁₀ / 13 L O₂ = 4.08 L C₄H₁₀ used up

Liters of C₄H₁₀ remaining = 4.35 L - 4.08 L = 0.270 C₄H₁₀ L remaining