Consider the following reaction equation
C2H4(g) + 3O2(g) \(\to\) 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
- 64.2dm3
-
33.6dm3
- 11.2dm3
- 3.73dm3
The correct answer is: B
Explanation
C2H4(g) + 3O2(g) \(\to\) 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 \(\frac{14 \times 3 \times 22.4dm^3}{28}\)
= 33.6dm3