Need help with a chemistry question!

The question is:
If we roast 95g of iron(II) sulfide, how many grams of oxygen will be needed, with how many grams will the solid material weigh more or less?
_FeS + _O2 => _Fe2O3 + _SO2 (this was given besides the question)

Thanks!!

2 FeS + 2.5 O2 => Fe2O3 + 2 SO2
2 x (2 FeS + 2.5 O2 => Fe2O3 + 2 SO2) =
4 FeS + 7 O2 => 2 Fe2O3 + 4 SO2


95 g given / 87.9 g/mole FeS = 1.08077 moles FeS x 7 moles O2 / 4 moles FeS (molar ratio) =
1.8913 moles of O2 x 32.00 g/mole for O2 = 60.523 g O2
or 61g to 2 S.F.

95 g given FeS / 87.9 g/mole FeS = 1.08077 moles FeS x 2 moles Fe2O3 / 4 moles FeS (molar ratio)
= 0.540385 moles of Fe2O3
0.540385 moles of Fe2O3 x 159.7 g/mole Fe2O3 = 86.299 g Fe2O3
86 g to 2 S.F.