# Question #39c68

Aug 22, 2016

$0.50 {\text{ dm}}^{3}$ or $0.50 \text{ L}$.

#### Explanation:

This is based on the balanced reaction:

$H C \equiv C H \left(g\right) + 2 {H}_{2} \left(g\right)$
$\rightarrow {H}_{3} C - C {H}_{3} \left(g\right)$.

Thus one mole of ethyne (${\text{C"_2"H}}_{2}$) requires two moles of hydrogen. Given equiv temperature and pressure for a reaction between (essentially ideal) gases, the same volume ratio applies. Thus $0.25 \cdot {\mathrm{dm}}^{3}$ acetylene clearly requires $0.50 \cdot {\mathrm{dm}}^{3}$ dihydrogen gas for stoichiometric equivalence.

Note that $1 \text{ dm"^3=(10^-1" m")^3 =10^-3" m"^3=1" L}$.