What is the mass in grams of the excess Fe₂O₃ remaining after the partial reaction of 201 g Fe₂O₃ with 64.7 g Al? Give your answer to three significant figures.

Answer 1

Step-by-step explanation

I'm assuming that the reaction is:

2 Al+Fe2​O3​→2 Fe+Al2​O3​

Calculate the number of moles of each reactant, using MAl​=26.982 g/mol, MFe2​O3​​=2⋅55.845+3⋅15.999=159.687 g/mol:

n=Mm​

nAl​=26.982 g/mol64.7 g​=2.40 mol

nFe2​O3​​=159.687 g/mol201 g​=1.26 mol

Assume that Al is the limiting reactant and calculate how many moles of Fe2O3 are stoichiometrically required to consume the amount of Al present:

nFe2​O2​, stoich.​=2.40 molAl​⋅2 molAl​1 molFe2​O3​​​=1.20 molFe2​O3​​

Since the stoichiometric amount of Fe2O3 is less than the actual amount, it's indeed in excess and Al is the limiting reactant. Calculate the number of moles of Fe2O3 in excess:

nFe2​O3​, exc.​=nFe2​O3​, actual​−nFe2​O3​, stoich.​

nFe2​O3​, exc.​=1.26 mol−1.20 mol=0.06 mol

Calculate the mass of Fe2O3 in excess:

nFe2​O3​, exc.​=0.06 mol⋅159.687 g/mol=9.5 g

Yes, Al is the limiting reactant and Fe2O3 is in excess by 9.5 g.