For the balanced equation shown below, if the reaction of 77.0 gms of ${\mathrm{CaCN}}_2$ produces 27.1 gms of ${\mathrm{NH}}_3.$ What is the percent yield?

${\mathrm{CaCN}}_2+3{\mathrm{H}}_2\mathrm{O}\to {\mathrm{CaCO}}_3+2{\mathrm{NH}}_3$

The given chemical reaction is ${\mathrm{CaCN}}_2+3{\mathrm{H}}_2\mathrm{O}\to {\mathrm{CaCO}}_3+2{\mathrm{NH}}_3$

1 mole of ${\mathrm{CaCN}}_2\to 2 \mathrm{mole} \mathrm{of} {\mathrm{NH}}_3$

80 g of   ${\mathrm{CaCN}}_2\to 2\times 17\mathrm{g} \mathrm{of} {\mathrm{NH}}_3$

The amount of Ammonia should be yield from 77g of${\mathrm{CaCN}}_2 \mathrm{is}$

$\frac{2\times 17 \mathrm{g}}{80\mathrm{g}}\times 77.0g=32.725 g$

Now, percentage yield of Ammonia in the reaction is

$\frac{\mathrm{Actual} \mathrm{yield}}{\mathrm{Theoritical} \mathrm{yield}}\times 100=\frac{27.1g}{32.725}\times 100=82.81=83\%$