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SOLUTION:

Conversion of units:

1 kg = 1,000 g
1 mol C7H6O3 = 138.12354 g C7H6O3
1 mol C4H6O3 = 1 mol C7H6O3
1 mol C4H6O3 = 102.09024 g C4H6O3

[(181 Kg C7H6O3)/1][(1,000 g)/(1 Kg)][(1 mol C7H6O3)/(138.12354 g
C7H6O3)][(1 mol C4H6O3)/(1 mol C7H6O3)][(102.09024 g C4H6O3)/(1 mol
C4H6O3)][(1 Kg)/(1,000 g) = 134 Kg C4H6O3

Since only about 143 Kg of C4H6O3 is needed, as much as 167 Kg of C4H6O3
is available, we can assume that C4H6O3 is present in excess, and
C7H6O3 is the determining reactant.

Conversion of units:

1 kg = 1,000 g
1 mol C7H6O3 = 138.12354 g C7H6O3
1 mol C9H8O4 = 1 mol C7H6O3
1 mol C9H8O4 = 180.16102 g C9H8O4

[(181 Kg C7H6O3)/1][(1,000 g)/(1 Kg)][(1 mol C7H6O3)/(138.12354 g
C7H6O3)][(1 mol C9H8O4)/(1 mol C7H6O3)][(180.16102 g C9H8O4)/(1 mol
C9H8O4)][(1 Kg)/(1,000 g) = 236 Kg C9H8O4

Answer: The theoretical yieldis 252
Kg.

% yield = [(actual yield)(100%)] / (theoretical yield)
% yield of C9H8O4 = [(173 Kg)(100%)] / (236 Kg)
% yield of C9H8O4 = 73.3%

Answer: the percentage yield is 73.3%.

** P/S: Please remember to give me a review later on once the answer is correct. I would be highly appreciate it. Thank you. **

Jul 24th, 2015

** P/S: Please remember to give me a
review later on once the answer is correct. I would be highly appreciate it.
Thank you. **

Jul 24th, 2015

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