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 b...

The balanced chemical equation is: C2H4(g) + 3O2(g) -> 2CO2(g) + 2H2O(g) From the equation, we can see that 1 mole of C2H4 produces 2 moles of CO2. Therefore, 0.05 moles of C2H4 will produce: 2 x 0.05 = 0.1 moles of CO2 Now we need to find the volume of 0.1 moles of CO2 at s.t.p. The molar volume of gas at s.t.p. is 22.4 dm3. So, the volume of 0.1 moles of CO2 at s.t.p. will be: 0.1 x 22.4 = 2.24 dm3 Therefore, the volume of CO2 produced at s.t.p when 0.05 moles of C2H4 was burnt in O2 is 2.24 dm3. Hence, the correct option is (b) 2.24 dm3.