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

  • A 64.2dm3
  • B 33.6dm3
  • C 11.2dm3
  • D 3.73dm3
waec 2016chemistry

The correct answer is B. 33.6dm3

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

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