Combining 0.201 mol of Fe_2O_3 with excess carbon produced 16.0 g of Fe. What is
ID: 558314 • Letter: C
Question
Combining 0.201 mol of Fe_2O_3 with excess carbon produced 16.0 g of Fe.What is the actual yield of iron in moles?
What was the theoretical yield of iron in moles?
What was the percent yield?
Map O Sapling Learning Combining 0.201 mol of FezO3 with excess carbon produced 16.0 g of Fe. Fe203 + 3C 2Fe+3CO What is the actual yield of iron in moles? Number mol What was the theoretical yield of iron in moles? Number mol What was the percent yield? Number Previous Give Up & View Solution Check AnswerNext Exit ? Hint
Explanation / Answer
1)
Molar mass of Fe = 55.85 g/mol
mass of Fe = 16.0 g
we have below equation to be used:
number of mol of Fe,
n = mass of Fe/molar mass of Fe
=(16.0 g)/(55.85 g/mol)
= 0.287 mol
Answer: 0.287 mol
2)
from reaction,
moles of Fe formed = 2*moles of Fe2O3
= 2*0.201 moles
= 0.402 mol
Answer: 0.402 mol
3)
% yield = actual yield * 100 / theoretical yield
= 0.287*100/0.402
= 71.4 %
Answer: 71.4 %