Molar mass
O2 = 31.99 g/mol
Fe = 55.8 g/mol
Fe2O3 = 159.68 g/mol
Identifying excess reagent and the limiting of the reaction :
3 O2 + 4 Fe = 2 Fe2O3
3 x 31.99 g O2 ----------- 2 x 159.68 g Fe2O3
?? g O2 ----------------- 250 g Fe2O3
250 x 3 x 31.99 / 2 x 159.68 =
23992.5 / 319.36 => 75.12 g of O2 iron is the limiting reactant
4 x 55.8 g Fe ------------ 2 x 159.68 g Fe2O3
175 g Fe ----------------- ??
2 x 159.68 x 175 / 4 x 55.8 =
55888 / 223,2 => 250.394 g of Fe2O3 , Fe is the excess reagent
therefore 75.12 g of O2
hope this helps!
