Chinenye G. answered 11/03/21
Chemistry, Biology and Statistics tutor
With this problem you can use substance to coefficient ratio method to determining which one is the limiting reactant versus the excess reactant. The limiting reactant will have the smaller ratio. The excess will have the bigger or greater ratio.
You are told that you have 100g of both reactants. Take 100g of each reactant and take it to their moles by their atomic weights.
FeS: at weight: 87.91g/mol
100g * 1mol/87.91g grams will cancel top to bottom and you will have:. 1.1375 mol FeS
1.1375 mol/4 mol = 0.283375
Now find the same for diatomic oxygen O2:
At weight O2:. 32g/mol
100g* 1mol/32g. = 3.125 mol
3.125mol/7mol = 0.4464
So O2 is the excess,. While FeS is the limiting reactant because it has the smaller ratio number.
So now proceed the reaction with the limiting reactant:
According to the balanced equation,
4FeS. ==> 2Fe2O3
1.1375 mol ==> XFe2O3
Cross multiply and solve for X
4X = 2(1.1375)
X = 0.56875 mol
Now take this to grams of Fe2O3:
Fe2O3 at weight: 112.13g/mol
0.56875 mol* 112.13g/mol
and moles will cancel top to bottom and you will have
63.774 g or 63.8 g (for 3 sig figs) of Fe2O3.