Which of the following statements is correct when 2.0mol of fes(s) reacts with 4.0mol of o2(g) ?

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.

4 FeS + 7 O2 → 2 Fe2O3 + 4 SO2 ... balanced equation

One easy way to find the limiting reactant is to divide the moles of each reactant by the corresponding coefficient in the balanced equation. For this problem, we have...

FeS: 100 g FeS x 1 mol FeS / 87.9 g = 1.14 mols FeS (÷4 -> -.284)

O2: 100 g O2 x 1 mol O2 / 32 g = 3.125 mols O2 (÷7 ->0.446)

FeS is LIMITING since 0.284 is less than 0.446.

Now we use the moles of limiting reactant to find the theoretical yield of Fe2O3:

1.14 mols FeS x 2 mols Fe2O3 / 4 mols FeS = 0.57 mols Fe2O3

If you want the mass of Fe2O3 then it is 0.57 mols x 160 g/mol = 91.2 g Fe2O3

The reactant in excess must be the other reactant, O2.