ron reacts with oxygen at high temperatures to form iron(III) oxide. 4Fe(s)+3O2(g)⟶2Fe2O3(s) Suppose 13.0 g of iron (Fe) is reacted with 19.7 g of oxygen (O2).

Question

What is the limiting reagent?

Calculate the theoretical yield of iron(III) oxide

The reaction produces 8.59 g of iron(III) oxide. What is the percent yield of the reaction?

Answer 1

4Fe(s) + 3O₂(g) ==> 2Fe₂O₃(s) ... balanced equation

One easy way to find the limiting reactant is to simply divide the moles of each reactant by the corresponding coefficient in the balanced equation.

For Fe: 13.0 g Fe x 1 mol Fe / 55.9 = 0.233 mols Fe (÷4->0.058) =LIMITING REAGENT

For O2: 19.7 g O₂ x 1 mol O₂ / 32 g = 0.616 mols O₂ (÷3->0.21)

Since 0.058 is less than 0.21, Fe is LIMITING and will determine the amount of product formed

Theoretical yield of Fe₂O₃

0.233 mols Fe x 2 mols Fe₂O₃ / 4 mols Fe x 160 g Fe₂O₃ / mol = 18.6 g Fe₂O₃ = theoretical yield

Percent yield

% yield = actual yield / theoretical yield (x100%)

% yield = 8.59 g / 18.6 g (x100%) = 46.2% yield