You need 4 mols of CO for every 1 mol of Fe
4 * .03820 = .153
so we need .153 mols of CO
BUT
we only have .120 mols of CO so the CO limits or controls the reaction
so we actually use .120/4 = .0300 mols of Fe3O4
that means we have .0300 *3 mols of Fe = .0900 mols Fe on each side
so
.0900 * 55.8 grams/mol = 5.02 grams Fe
Consider this reaction: Fe3O4(s)+4CO(g) --> 4CO2(g)+3Fe(s). If you begin with 0.03820 moles of iron oxide and 0.120 moles of carbon monoxide, how many grams of iron can be formed? Indicate which reactant is the limiting reactant.
0.03820 x 3/1= 0.1146 moles x 55.8 g
0.120 x 1/4= 0.03 moles x 55.8 g
9.8 g
3 answers
correction to first line Fe3O4 not Fe
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You need 4 mols of CO for every 1 mol of Fe3O4
4 * .03820 = .153
so we need .153 mols of CO
BUT
we only have .120 mols of CO so the CO limits or controls the reaction
so we actually use .120/4 = .0300 mols of Fe3O4
that means we have .0300 *3 mols of Fe = .0900 mols Fe on each side
so
.0900 * 55.8 grams/mol = 5.02 grams Fe
----------------------
You need 4 mols of CO for every 1 mol of Fe3O4
4 * .03820 = .153
so we need .153 mols of CO
BUT
we only have .120 mols of CO so the CO limits or controls the reaction
so we actually use .120/4 = .0300 mols of Fe3O4
that means we have .0300 *3 mols of Fe = .0900 mols Fe on each side
so
.0900 * 55.8 grams/mol = 5.02 grams Fe
I agree with 5.02 g Fe produced.