Question

If 5g of H2 are reacted with excess CO, how many grams of CH3OH are produced, based on a yield of 86%?

Answer 1

Answer: 34.4 g

Explanation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

$\text{Number of moles of hydrogen}=\frac{5g}{2g/mol}=2.5moles$

As $CO$ is in excess, $H_2$ is the limiting reagent and thus it will limit the formation of products.

$CO+2H_2\rightarrow CH_3OH$

According to stoichiometry:

2 moles of hydrogen produce = 1 mole of $CH_3OH$

2.5 moles of hydrogen produce = $\frac{1}{2}\times 2.5=1.25 moles$ of $CH_3OH$

Mass of $CH_3OH=moles\times {\text {Molar mass}}=1.25\times 32=40g$

But as % yield is 86%, mass of $CH_3OH$ produded is $\frac{86}{100}\times 40=34.4g$

Thus 34.4 g of $CH_3OH$ is produced.