Question

The following reaction is used to obtain iron from iron ore:

Fe2O3(s)+3CO(g) ? 2Fe(s)+3CO2(g)

The reaction of 172 g of Fe2O3 with 84.2 g of CO produces 71.8 g of Fe.

Calculate the theoretical yield of solid iron.

Express the mass in grams to three significant figures.

Answer 1

The reaction is,

Fe2O3(s)+3CO(g) → 2Fe(s)+3CO2(g)

According to the stoichiometry of the reaction 1 mole of Fe2O3 reacts with 3 moles of CO to produce 2 moles of Fe

Mass of Fe2O3 reacting = 172 g

Mass of CO = 84.2 g

Mass of Fe produced = 71.8 g

We know that,

Molar Mass of Fe2O3 = 159.7 g / mol

Molar Mass of CO = 28 g / mol

Atomic Mass of Fe = 55.845 g / mol

=> Moles of Fe2O3 = 172 / 159.7 = 1.077 moles

Moles ofCO = 84.2 / 28 = 3.007

Moles of Fe = 71.8 / 55.845 = 1.286 moles

3 moles of CO reacts with 1 mole of Fe2O3

=> 3.007 moles of Fe reacts with (1/3) x 3.007 = 1.002 moles of Fe2O3. After this there will be no CO left to react.

Therefore Fe2O3 is in excess

Now,

3 moles of CO produces 2 moles of Fe (Theoretically)

=> 3.007 moles of CO produces (2/3) x 3.007 = 2.005 moles of Fe

=> Theoretical moles of Fe = 2.005

=> Theoretical Mass = 2.005 x 55.845 = 111.95 g

% yield = (71.8 / 111.95) x 100 = 64.14 %