Question

ch 1-8

22.) The following reaction takes place at a certain elevated temperature:

Fe2O3(s) + 3CO(g) → 2Fe(l) + 3CO2(s)

What is the percent yield of iron if 57.1 g Fe2O3 in excess CO produces 17.9 g Fe? The M.W. of Fe2O3 is 159.7 g/mol and the M.W. of CO is 28.01 g/mol.

Recall that the percent yield is the actual yield divided by the theoretical yield times 100%. Additional help is given in the feedback.

Procedure: g Fe2O3 --> mol Fe2O3 --> mol Fe --> g Fe (This is the theoretical yield. Note M.W. CO is not used.) Take actual yield, 17.9, and divide by the theoretical yield and then multiply by 100 (to change to percent yield).

Answer 1

The balanced chemical equation for the reaction is given as

Fe₂O₃ (s) + 3 CO (g) -------> 2 Fe (s) + 3 CO₂ (g)

As per the stoichiometric equation,

1 mole Fe₂O₃ = 2 moles Fe.

CO is in excess; hence, Fe₂O₃ is the limiting reactant and the yield of the product is decided by the amount of Fe₂O₃ present in mole(s).

The M.W. of Fe₂O₃ is given as 159.7 g/mol.

Mole(s) of Fe₂O₃ corresponding to 57.1 g Fe₂O₃ = (57.1 g)*(1 mole/159.7 g) = 0.3575 mole.

Mole(s) of Fe produced theoretically = (0.3575 mole Fe₂O₃)*(2 moles Fe/1 mole Fe₂O₃) = 0.7150 mole.

The atomic mass of Fe = 55.845 g/mol.

The theoretical yield of Fe = (0.7150 mole)*(55.845 g/mol) = 39.9292 g.

The actual yield of Fe is 17.9 g; therefore, the percent yield is (actual yield)/(theoretical yield)*100 = (17.9 g)/(39.9292 g)*100 = 44.8293% ≈ 44.83% (ans).