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?

Question

contestada

Respuesta :

Mergus Mergus · Answer 1 · 2019-12-12T05:54:51+01:00

Answer:

[tex]\%\ yield =61.15\ \%[/tex]

Explanation:

1 mole of STP has a volume of 22.4 L

So,

Number of the moles of [tex]CO[/tex] which are being fed per minute = [tex]\frac{23.3}{22.4}\ moles/minute=1.0402\ moles/minute[/tex]

Also, Number of the moles of [tex]H_2[/tex] which are being fed per minute = [tex]\frac{10.7}{22.4}\ moles/minute=0.4777\ moles/minute[/tex]

According to the reaction shown below:-

[tex]CO+2H_2\rightarrow CH_3OH[/tex]

1 mole of [tex]CO[/tex] reacts with 2 moles of [tex]H_2[/tex]

1.0402 mole of [tex]CO[/tex] reacts with 2*1.0402 moles of [tex]H_2[/tex]

Moles of [tex]H_2[/tex] = 2.0804 moles

Thus, [tex]H_2[/tex] is the limiting reagent.

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

2 moles of [tex]H_2[/tex] on reaction forms 1 mole of [tex]CH_3OH[/tex]

1 mole of [tex]H_2[/tex] on reaction forms 1/2 mole of [tex]CH_3OH[/tex]

Also,

0.4777 mole of [tex]H_2[/tex] on reaction forms [tex]\frac{1}{2}\times 0.4777[/tex] mole of [tex]CH_3OH[/tex]

Moles of [tex]CH_3OH[/tex] = 0.23885 moles

Also, Molar mass of [tex]CH_3OH[/tex] = 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:-

[tex]\%\ yield =\frac {Experimental\ yield}{Theoretical\ yield}\times 100[/tex]

Given , Values from the question:-

Theoretical yield = 7.6528 g

Experimental yield = 4.68 g

Applying the values in the above expression as:-

[tex]\%\ yield =\frac{4.68}{7.6528}\times 100[/tex]

[tex]\%\ yield =61.15\ \%[/tex]