Respuesta :
Explanation:
It is know that, 1 mole of a gas at STP occupies a volume of 22. 4 L. Hence, calculate the number of moles of CO as follows.
No. of moles of CO being fed per minute = [tex]\frac{25.0 L}{22.4 L}[/tex] per minute
= 1.116 moles per minute
No. of moles of [tex]H_{2}[/tex] being fed per minute = [tex]\frac{16.0 L}{22.4 L}[/tex] per minute
= 0.714 moles per minute
As, the balanced reaction equation is as follows.
[tex]CO + 2H_{2} \rightarrow CH_{3}OH[/tex]
Here, [tex]H_{2}[/tex] is the limiting reagent. Since, 2 moles [tex]H_{2}[/tex] is equivalent to 1 mol methanol produced .
Or, 0.71 moles of [tex]H_{2}[/tex] contains [tex](\frac{1}{2}) \times 0.71 moles[/tex]
= 0.357 moles of methanol.
Therefore, theoretical yield of methanol = 0.357 moles per minute.
Observed yield of methanol = 5.30 gram per minute
= [tex]\frac{5.30 gram}{32.04 g/mol}[/tex]
= 0.165 moles per minute
Nowe, we will calculate the percentage yield as follows.
% yield = [tex]\frac{\text{observed yield}}{\text{theoretical yield}} \times 100[/tex]
= [tex]\frac{0.165 moles/min}{0.357 moles/min} \times 100[/tex]
= 46.32%
Thus, we can conclude that percent yield of the reaction is 46.32%.
