# How many grams of  CH_3OH can be made from 23.12 g of CO and 8.323 of H_2?

Feb 11, 2017

We need (i) a stoichiometric equation........

#### Explanation:

$C O + 2 {H}_{2} \rightarrow C {H}_{3} O H$

And (ii) equivalent quantities of carbon monoxide and dihydrogen.

$\text{Moles of carbon monoxide}$ $=$ $\frac{23.12 \cdot g}{28.01 \cdot g \cdot m o {l}^{-} 1} = 0.825 \cdot m o l$.

$\text{Moles of dihydrogen}$ $=$ $\frac{8.323 \cdot g}{2.016 \cdot g \cdot m o {l}^{-} 1} = 2.016 \cdot m o l$.

Clearly, the dihydrogen is in stoichiometric excess, and at most we can make $0.825 \cdot m o l$ of methyl alcohol.

And thus, maximum quantity of methyl alcohol,

$= 0.825 \cdot m o l \times 32.04 \cdot g \cdot m o {l}^{-} 1 = 26.4 \cdot g$.

The mixture of carbon monoxide and dihydrogen is called syngas. This is usually used directly without conversion to methyl alcohol.