Question

Methanol (CH3OH) can be produced by the following reaction: CO(g) + 2H2(g) → CH3OH(g) Hydrogen at STP flows into a reactor at a rate of 10.7 L/min. Carbon monoxide at STP flows into the reactor at a rate of 23.3 L/min. If 4.68 g of methanol is produced per minute, what is the percent yield of the reaction?

Answer 1

`\%\ yield =61.15\ \%`

Step-by-step explanation:

1 mole of STP has a volume of 22.4 L

So,

Number of the moles of
`CO` which are being fed per minute =
`(23.3)/(22.4)\ moles/minute=1.0402\ moles/minute`

Also, Number of the moles of
`H_2` which are being fed per minute =
`(10.7)/(22.4)\ moles/minute=0.4777\ moles/minute`

According to the reaction shown below:-

`CO+2H_2\rightarrow CH_3OH`

1 mole of
`CO` reacts with 2 moles of
`H_2`

1.0402 mole of
`CO` reacts with 2*1.0402 moles of
`H_2`

Moles of
`H_2` = 2.0804 moles

Thus,
`H_2` is the limiting reagent.

The formation of the product is governed by the limiting reagent. So,

2 moles of
`H_2` on reaction forms 1 mole of
`CH_3OH`

1 mole of
`H_2` on reaction forms 1/2 mole of
`CH_3OH`

Also,

0.4777 mole of
`H_2` on reaction forms
`(1)/(2)* 0.4777` mole of
`CH_3OH`

Moles of
`CH_3OH` = 0.23885 moles

Also, Molar mass of
`CH_3OH` = 32.04 g/mol

Mass = Moles*Molar mass = 0.23885*32.04 g = 7.6528 g

The expression for the calculation of the percentage yield for a chemical reaction is shown below as:-

`\%\ yield =\frac {Experimental\ yield}{Theoretical\ yield}* 100`

Given , Values from the question:-

Theoretical yield = 7.6528 g

Experimental yield = 4.68 g

Applying the values in the above expression as:-

`\%\ yield =(4.68)/(7.6528)* 100`

`\%\ yield =61.15\ \%`