Consider the following reaction equation

C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(l).

The volume of oxygen at s.t.p that will be required to burn 14g of ethene is [C2H4 = 28; Molar volume of gas at s.t.p = 22.4dm³]

A. 64.2dm³
B. 33.6dm³
C. 11.2dm³
D. 3.73dm³

Correct Answer: Option B

B. 33.6dm³

Explanation

C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(l)

1 mole of ethene required 3 moles of O2
28g of ethene required 3 x 22.4dm³ of O2

14g will require 14×3×22.4dm³/28

= 33.6dm3