Calculate the theoretical yield of iron(III) oxide ( Fe2O3 ) And The reaction produces 8.36 g of Fe2O3. What is the percent yield of the reaction?

Question

Iron reacts with oxygen at high temperatures to form iron(III) oxide.

4Fe(s)+3O2(g)⟶2Fe2O3(s)

Suppose 18.0 g of iron (Fe) is reacted with 20.7 g of oxygen 

Answer 1

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

First, find the limiting reactant. Just divide moles of each reactant by the corresponding coefficient in the balanced equation.

For Fe we have ... 18.0 g x 1 mol Fe / 55.9 g = 0.322 mols Fe (÷4->0.08)

For O2 we have ... 20.7 g O2 x 1 mol O2 / 32 g = 0.647 mols O2 (÷3->0.2)

Since 0.08 is less than 0.2, Fe is limiting

Next, use MOLES of Fe to find the mols of Fe2O3 that can be formed (theoretical yield)

Theoretical yield = 0.322 mols Fe x 2 mols Fe2O3 / 4 mols Fe x 159.7 g Fe2O3/ mol = 25.7 g Fe2O3

Finally, calculate % yield

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

% yield = 8.36 g / 25.7 g (x100%) = 32.5% yield