Consider the following reaction: 2A+B -> 3C+D
3.38 mol of A and 2.00 mol of B react to form 4.00 mol of C. What is the percent yield of this reaction?
a. 21.1%
b. 52.6%
c. 39.4%
d. 78.9%
e. 100.%
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To calculate the percent yield, we need to compare the actual yield (# of moles of C produced) to the theoretical yield (the amount of C that would be produced if the reaction went to completion).
From the balanced equation, we can see that the stoichiometry is 2:1 between A and C. So if we start with 3.38 mol of A, the theoretical yield of C would be:
(3.38 mol A) x (3 mol C/2 mol A) = 5.07 mol C
Therefore, the theoretical yield of C is 5.07 mol.
The percent yield is then calculated as:
(actual yield / theoretical yield) x 100%
In this case, the actual yield is given as 4.00 mol, and the theoretical yield is 5.07 mol.
(percent yield) = (4.00 mol / 5.07 mol) x 100% = 78.9%
So the answer is option d. 78.9%.
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