Chemistry, Stoichiometry Problem Help

Magnetite, Fe3O4, can be converted into metallic iron by heating with carbon monoxide as represented by this equation.
Fe3O4(s) + 4 CO(g) ---> 3 Fe(s) + 4 CO 2(g)
How many kilograms of Fe3O4 must be processed in this way to obtain 5.00 kg of iron if the process is 85% efficient?
Can someone show me how to find kilograms of Fe3O4? I'm not sure if you would convert Fe into kg and how? And what does 85% efficient mean?

(1) Calculate the number of moles of Fe that are in 5.00 kg

(5.00 kg Fe)/(55.85 kg/kmol) = 0.08953 kmol Fe

(2) Calculate the number of moles of Fe3O4 needed according to the reaction

(0.08953 kmol Fe)*((1 mol Fe3O4)/(3 mol Fe)) = 0.02984 kmol of Fe3O4

(3) Calculate the mass of that many moles of Fe3O4

(0.02984 kmol of Fe3O4)*(231.55 kg/kmol) = 6.9099 kg of Fe3O4

(4) Now we account for the efficiency by dividing the weigh of Fe3O4 by the efficiency as a decimal:

(6.9099 kg of Fe3O4)/(0.85) = 8.129 kg of Fe3O4 (8.13 kg to three sig figs)

Find the number of moles of iron.
Fe = 5000 grams / 56 grams per mole = 89.29 moles

Find the number of moles of Fe3O4.

1 mole of Fe3O4 is needed to produce 3 moles of Fe
x mole of Fe3O4 is needed to produce 89.29 moles of Fe

3x = 89.29
x = 29.76 moles of Fe3O4 are needed to produce 89.29 moles of Fe.

Now here's the sticky part. Your process is only 85% efficient. That means that we need more to get the job done.

85% of Fe3O4 is 29.76 moles
100% of Fe3O4 is x moles

85/100 = 29.76/x
85 x = 100 * 29.76
x = 100 * 29.76 / 85
x = 35.0 moles are required.

Now we need to convert 35 moles into kg.

1 mole of Fe3O4 = 3*56 + 4*16 = 232 grams
35 mole = 232 * 35 = 8120 grams

8120 grams * [ 1 kg/1000 grams ] = 8.12 kg.