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
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
-
A.
64.2dm3 -
B.
33.6dm3 -
C.
11.2dm3 -
D.
3.73dm3
Correct Answer: Option 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