1. The reaction of a mixture of CO and H2 is predicted to produce 554 g of CH3OH
ID: 1006583 • Letter: 1
Question
1.
The reaction of a mixture of CO and H2 is predicted to produce 554 g of CH3OH.
CO(g) + 2 H2(g) CH3OH()
If only 465 g of CH3OH is actually produced, what is the percent yield of the compound?
____________%
2.
Black smokers are found in the depths of the oceans. Thinking that the conditions in these smokers might be conducive to the formation of organic compounds, two chemists in Germany found the following reaction could occur in similar conditions.
2CH3SH + CO CH3COSCH3 + H2S
If you begin with 43.7 g of CH3SH and excess CO, what is the theoretical yield of CH3COSCH3? If you isolate 23.8 g of CH3COSCH3, what is the percent yield of CH3COSCH3?
_________g CH3COSCH3
_________%
Explanation / Answer
1.
Percent yield = (Amount of product formed / Theoretically meximum amount that can form) X 100
Percent yield = (465 / 554 ) X 100 = 83.9%
2.
2CH3SH + CO CH3COSCH3 + H2S
(i) As per the stiochiometry of the rection two moles or 96g of CH3SH will form one mole or 90g (Theoretical yield) of CH3COSCH3.
( 96g of CH3SH = 90g of CH3COSCH3)
1g of CH3SH will for 90.0 / 96.0 = 0.94g of CH3COSCH3
Therefore 43.7 g of CH3SH will form 0.94 X 43.7 = 41.1g
(ii) %yield = (Practical yield / theoretical yield ) X 100 = (23.8 / 41.1) X 100 = 57.9 %
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