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

4 Fe(s) + 3 O2(g) → 2 Fe2O3(s)
4(55.85) Fe(s) + 3(2*16) O2(g) → 2(2*55.85 + 3*16) Fe2O3(s)

a) x = 8*2(2*55.85 + 3*16)/( 4(55.85)) lb

b)x = 8*3(2*16)/( 4(55.85)) lb