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