Consider the following reaction equation:

C2H4(g) + 3O2(g) \(\to\) 2CO2(g) + 2H2O(g)

The volume of CO2(g) produced at s.t.p when 0.05 moles of C2H4(g) was burnt in O2(g) is

[Molar Volume of gas = 22.4dm3

  • A 1.12dm3
  • B 2.24dm3
  • C 3.72dm3
  • D 4.48dm3
waec 2015chemistry

The correct answer is B. 2.24dm3

\(\frac{\text{volume}}{\text{molar volume}}\) = 0.05

1 mole of ethane produces 2 moles of carbon(iv)oxide

∴ 0.05  mole of ethane will produce x moles of carbon(iv)oxide

= \(\frac{\text{volume}}{22.4}\) = 0.05 X 2 X 22.4 =  2.24dm\(_3\)

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