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

Step-by-step explanation:

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

`\text{Number of moles of hydrogen}=(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 =
`(1)/(2)* 2.5=1.25 moles` of
`CH_3OH`

Mass of
`CH_3OH=moles* {\text {Molar mass}}=1.25* 32=40g`

But as % yield is 86%, mass of
`CH_3OH` produded is
`(86)/(100)* 40=34.4g`

Thus 34.4 g of
`CH_3OH` is produced.