# Question #6e7c9

Dec 7, 2014

The answers are ${V}_{{H}_{2}} = 43.1 L$ and ${V}_{C O} = 21.5 L$.

Starting from the ideal gas law $P V = n R T$ and from the chemical reaction's balanced equation

$C O \left(g\right) + 2 {H}_{2} \left(g\right) \to C {H}_{3} O H \left(g\right)$, one can see that we get a $1 : 1$ mole ratio for $C O$ and $C {H}_{3} O H$, and a $2 : 1$ mole ratio for${H}_{2}$ and $C {H}_{3} O H$.

The number of moles of $C {H}_{3} O H$ can be calculated from

${n}_{C {H}_{3} O H} = {m}_{C {H}_{3} O H} / \left(32 \frac{g}{m o l}\right) = \frac{25.8 g}{32 \frac{g}{m o l}} = 0.81$ moles.
Subsequently, the number of moles for $C O$ and ${H}_{2}$ are

${n}_{C O} = 0.81$ moles, and ${n}_{{H}_{2}} = 1.62$ moles.

Therefore, the volume of ${H}_{2}$ required for the reaction to take place is

${V}_{{H}_{2}} = \frac{{n}_{{H}_{2}} R T}{P} = \frac{1.62 \cdot 0.082 \cdot 319.15}{\frac{748}{760}} = 43.1$ L ( I've converted pressure into $a t m$ anddegrees Celsius into $K$).

The volume of $C O$ required will be

${V}_{C O} = \frac{{n}_{C O} R T}{P} = \frac{0.81 \cdot 0.082 \cdot 319.15}{\frac{748}{760}} = 21.5$ L, which is approximately half of the volume of ${H}_{2}$ determined (half the number of moles -> half the volume);