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!

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.

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

4.0 g C

M(C) = 12 g/mol

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

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.

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

Actual yield Fe is 15 g.

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

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

% yield = 53.8%

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