# Question #9140f

37.5 litres of ${C}_{2} {H}_{2}$ is required and 37.5 litres of water vapour is produced.
$2 {C}_{2} {H}_{2 \left(g\right)} + 5 {O}_{2 \left(g\right)} \rightarrow 4 C {O}_{2 \left(g\right)} + 2 {H}_{2} {O}_{\left(g\right)}$
From the equation you can see that if 75.0L of $C {O}_{2}$ is produced then 75/2 = 37.5L of ${C}_{2} {H}_{2}$ is required since 1 mole ${C}_{2} {H}_{2}$ produces 2 mole $C {O}_{2}$.
2 moles ${C}_{2} {H}_{2}$ produces 2 moles ${H}_{2} O$.
So 37.5L ${C}_{2} {H}_{2}$ produces 37.5L ${H}_{2} O$ vapour.