Combining 0.313 mol of Fe2O3 with excess carbon produced 10.7 g of Fe. Fe2O3+3C
ID: 927363 • Letter: C
Question
Combining 0.313 mol of Fe2O3 with excess carbon produced 10.7 g of Fe.
Fe2O3+3C -----> 2Fe +3CO
What is the actual yield of iron in moles?
What was the theoretical yield of iron in moles?
What was the percent yield?
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Mg+Cu (NO3) ---> Mg(NO3)2 +Cu
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Explanation / Answer
Combining 0.313 mol of Fe2O3 with excess carbon produced 10.7 g of Fe.
Fe2O3+3C -----> 2Fe +3CO
What is the actual yield of iron in moles?
yield = mass of Fe change to mol
mol = mass/MW = 10.7/55.845 = 0.19160 mol of Fe
What was the theoretical yield of iron in moles?
theoretical yield is
0.313 mol of Fe2O3 shoudl produce 2*0.313 = 0.939 mol of Fe
What was the percent yield?
% yield = real/theoretical * 100 = 0.19160 /0.939 *100 = 20.40 %
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