Question

Iron is extracted from the mineral magnetite, Fe3O4:
Fe3O4 (s) + 2 C (s) -> 3 Fe (s) + 2 CO2 (s)

A 35.0 g sample of magnetite is reacted with 4.0 g of Carbon, and 15.0 g of Iron is produced. Determine the percentage yield of this reaction.

If you solve other of my questions as well. I will give 45 points!

Answer 1

Fe3O4 (s) + 2 C (s) -> 3 Fe (s) + 2 CO2 (s)

from equation 1 mol 2 mol

Given 35.0 g Fe3O4 and 4.0 g C , we need to convert them into moles.

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35.0 g Fe3O4

M(Fe3O4) = 55.8 g/mol

35.0 g Fe3O4 * 1 mol Fe2O3/55.8 g Fe2O3 = 0.6272 mol Fe2O3

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4.0 g C

M(C) = 12 g/mol

4.0 g C* 1mol C/12 g C = 0.3333 mol C

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Fe3O4 (s) + 2 C (s) -> 3 Fe (s) + 2 CO2 (s)

from equation 1 mol 2 mol

given 0.6272 mol 0.3333 mol

We can see that Fe3O4 is an excess reactant,

so to find the amount of Fe we are going to use carbon.

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Fe3O4 (s) + 2 C (s) -> 3 Fe (s) + 2 CO2 (s)

from equation 2 mol 3 mol

given 0.3333 mol x mol

x=0.3333*3/2= 0.500 mol Fe is theoretical yield

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Actual yield Fe is 15 g.

15 g Fe* 1 mol/55.8 g = 0.2688 mol Fe is actual yield

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% yield = (actual yield/ theoretical yield)* 100% = (0.2688/0.500)*100%=53.8%

% yield = 53.8%