Question

Iron reacts with oxygen gas to produce iron (III) oxide. If there are 52 grams of

iron present in this reaction how many grams of iron (III) oxide are produced
assuming a 100% yield? Also how many liters of oxygen are needed assuming
STP conditions?

Answer 1

Mass of Fe₂O₃ : 74.26 g

Volume of O₂ at STP = 15.624 L

Further explanation

Reaction

4Fe (s) + 3O₂ (g) ==> 2Fe₂O₃ (s)

mol Fe :

Ar Fe = 56 g/mol

`\tt (52)/(56)=0.93`

mol Fe₂O₃ :

`\tt (2)/(4)* 0.93=0.465`

mass Fe₂O₃ :

MW Fe₂O₃ : 159,69 g/mol

`\tt 0.465* 159,69=74.26~g`

mol O₂ :

`\tt (3)/(4)* 0.93=0.6975`

volume O₂ at STP(1 mol=22.4 L)

`\tt 0.6975* 22.4=15.624~L`