Map Sapling Learning Combining 0.237 mol of Fe203 with excess carbon produced 19

Question

Map Sapling Learning Combining 0.237 mol of Fe203 with excess carbon produced 19.6 g of Fe Fe,03 +3C 2Fe +3CO What is the actual yield of iron in moles? Number 351 mo What was the theoretical vield of iron in moles? Incorrect The theoretical yield of Fe is based on the given amount of Number 448 I Fe203. If all 0.237 mol of Fe2O3 reacted, then double that amount mol of Fe would have been produced according to the coefficients from the balanced chemical equation. What was the percent yield? Number 65

Explanation / Answer

1)

Molar mass of Fe = 55.85 g/mol


mass(Fe)= 19.6 g

use:
number of mol of Fe,
n = mass of Fe/molar mass of Fe
=(19.6 g)/(55.85 g/mol)
= 0.351 mol

2)
from reaction,
moles of Fe = 2*moles of Fe2O3
= 2*0.237 mol
= 0.474 mol
Answer: 0.474 mol

3)
% yield = actual yield * 100 / theoretical yield
= 0.351*100/0.474
= 74.1 %
Answer: 74.1 %

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