A student tried to recover all the gold metal from a 1.738 M solution of gold (III) sulfate by...

Percent yield is the percent ratio between the actual and theoretical yield of the process.

{eq}\rm \%\ yield=\dfrac{actual\ yield}{theoretical\ yield} \times 100\%{/eq}

The actual yield of the recovery is 150 pounds of gold (Au).

The theoretical yield is the maximum amount of gold that can be recovered. For this case, we have a 450 gallons of the gold solution which is equivalent to 1703.44 liters. Using the concentration, the theoretical yield is (molar of Au is 196.97 g/mol):

{eq}\rm theoretical\ yield=(Volume\ of\ solution)\times (Concentration\ of\ solution)\times (Molar\ mass\ of\ Au)\times (Conversion\ factor)\\ \rm theoretical\ yield=1703.44\ L\ \times \left( \dfrac{1.738\ mol\ Au}{L} \right) \times \left( \dfrac{196.97\ g\ Au}{mol\ Au}\right) \times \left( \dfrac{1\ lb}{453.59\ g} \right)\\ \rm theoretical\ yield= 1286\ lb\ Au{/eq}

The percent yield is:

{eq}\rm \%\ yield=\dfrac{actual\ yield}{theoretical\ yield} \times 100\%\\ \rm \%\ yield=\dfrac{150\ lb\ Au}{1286\ lb\ Au} \times 100\%\\ \boxed{\rm \%\ yield=11.7\%}{/eq}