Answer to Question #119507 in General Chemistry for Roudha

Question

4FeS2 + 11O2 → 2Fe2O3 + 8SO2
If 60 g of FeS2 reacts with 32 g of O2 . find the limiting reactant . find the mass of Fe2O3 that will be produced

Answer 1

The limiting reagent is the one that is totally consumed.

M (FeS2) = 119.98 g/mol

M (O2) = 32.00 g/mol

60 / 119.98 = 0.5 mol

32 / 32.00 = 1 mol (twice more!)

According to the equation, it takes more than twice moles of O2 to completely react with FeS2. So 1 mol O2 would be consumed fully. O2 - limiting reagent.

M (Fe2O3) = 159.69

According to the equation,

n (Fe2O3) = 2/11 n(O2) = 2/11 * 1 mol = 0.1818 mol

m (Fe2O3) = 0.1818 * 159.69 = 29 g