Answer to Question #162373 in Chemistry for Ava Brooke

Question

Using the following chemical equation how many liters of oxygen are consumed by burning 7.86 g of butane?

C₄H₁₀ + 5O₂ -> 3CO₂ + 4H₂O

Answer 1

C₄H₁₀ + 5O₂ → 3CO₂ + 4H₂O - unbalanced equation

Solution:

C₄H₁₀ - butane

The balanced chemical equation:

2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O

According to the equation above: n(C₄H₁₀)/2 = n(O₂)/13

Moles of C₄H₁₀ = Mass of C₄H₁₀ / Molar mass of C₄H₁₀

The molar mass of C₄H₁₀ is 58.12 g mol^-1.

Hence,

n(C₄H₁₀) = 7.86 g / 58.12 g mol^-1 = 0.1352 mol

n(O₂) = 13 × n(C₄H₁₀) / 2 = 13 × 0.1352 mol / 2 = 0.8788 mol

At STP, one mole of any gas occupies a volume of 22.4 L.

Thus 0.8788 mol of oxygen (O₂) occupies:

(0.8788 mol × 22.4 L) / 1 mol = 19.685 L of oxygen (O₂)

0.8788 mol of oxygen (O₂) occupies a volume equal to 19.69 L.

Answer: 19.69 liters of oxygen (O₂) are consumed.

However, if an unbalanced equation is used, 15.15 liters of oxygen (O₂) are consumed.