4 Fe(s) + 3 O2(g) 2 Fe2O3(s)?

Question

a)What is the maximum mass of iron(III)oxide that can be obtained from 3632 g (8.000 lb) of iron?
b)What mass of O2 is required to oxidize the iron to iron (III) oxide?

Answer 1

Atomic weights: Fe=55.845 O=16 Fe2O3= 143.69

8.000 lbFe x 1 lb-molFe/55.845 lbFe x 2 lbmolFe2O3/4 lbmolFe x 143.69 lbFe2O3/1 lbmolFe2O3 = (8.000)(2)(143.69)/(55.845)(4) = 10.29 lb Fe2O3

8.000 lbFe x 1 lb-mol Fe/55.845 lbFe x 3 lbmolO2/4 lbmolFe x 32 lbO2/1 lbmolO2 = (8.000)((3)(32)/(55.845)(4) = 55.01 lb O2