please help with question for chemistry

Question

Aluminum reacts with iron (II) oxide to produce aluminum oxide and iron. Find the percentage yield of iron if 85.0 g of iron (II) oxide reacts with 35.0 g of aluminum metal and 65 g of iron are produced.

Answer 1

First, write the balanced chemical equation:

2Al + 3FeO ==> Al2O3 + 3Fe (redox)

Next, determine which reactant is limiting:

moles FeO = 85.0 g x 1 mol/71.84 g = 1.18 moles 

moles Al = 35.0 g x 1 mol/26.98 g = 1.30 moles 

From the mole ratio of 3FeO : 2 Al, and the moles present, the FeO is in limiting supply 

Next, determine moles of Fe produced under these conditions:

1.18 moles FeO x 3 moles Fe/3 moles FeO = 1.18 moles of Fe should be produced

Determine mass of Fe equivalent to 1.18 moles:

1.18 moles Fe x 55.8 g/mole = 65.8 g Fe = theoretical yield

Percent yield = 65 g/65.8 g (x100%) = 98.8% yield