Question

A 52 g sample of iron reacts with 22 g of oxygen to form how many grams of iron oxide?

Express your answer using two significant figures.

Answer 1

Answer – We are given, mass of Fe = 52 g , mass of O₂ = 22 g

4 Fe + 3O₂ ------> 2 Fe₂O₃

First we need to calculate the moles of each reactant

Moles of Fe = 52 g / 55.845 g.mol^-1 = 0.931 moles

Moles of O₂ = 22 g / 31.998 g.mol^-1 = 0.687 moles

Now we need to calculate the limiting reactant

Moles of Fe₂O₃ from Fe

From the balanced equation

4 moles of Fe = 2 moles of Fe₂O₃

So, 0.931 moles of Fe = ? moles of Fe₂O₃

= 0.931 moles of Fe * 2 moles of Fe₂O₃ / 4 moles of Fe

= 0.466 moles of Fe₂O₃

Now moles of Fe₂O₃ from O₂

From the balanced equation

3 moles of O₂ = 2 moles of Fe₂O₃

So, 0.687 moles of O₂ = ? moles of Fe₂O₃

= 0.687 moles of O₂ * 2 moles of Fe₂O₃ / 3 moles of O₂

= 0.458 moles of Fe₂O₃

So limiting reactant is O₂, since the moles of Fe₂O₃ got lowest from the moles of O₂.

So moles of Fe₂O₃ = 0.458 moles

Mass of Fe₂O₃ = 0.458 moles * 159.687 g/mol

                         = 73 g

So, a 52 g sample of iron reacts with 22 g of oxygen to form 73 grams of iron oxide