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.

Question

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

Answer 1

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

Answer 2

correction to first line Fe3O4 not 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

Answer 3

I agree with 5.02 g Fe produced.