Answer
The percent yield of calcium nitride for this reaction is equal to 46.4 %
Work Step by Step
$ Ca $ : 40.08 g/mol
$$ \frac{1 \space mole \space Ca }{ 40.08 \space g \space Ca } \space and \space \frac{ 40.08 \space g \space Ca }{1 \space mole \space Ca }$$
$ Ca_3N_2 $ : ( 40.08 $\times$ 3 )+ ( 14.01 $\times$ 2 )= 148.26 g/mol
$$ \frac{1 \space mole \space Ca_3N_2 }{ 148.26 \space g \space Ca_3N_2 } \space and \space \frac{ 148.26 \space g \space Ca_3N_2 }{1 \space mole \space Ca_3N_2 }$$
$$ 56.6 \space g \space Ca \times \frac{1 \space mole \space Ca }{ 40.08 \space g \space Ca } \times \frac{ 1 \space mole \space Ca_3N_2 }{ 3 \space moles \space Ca } \times \frac{ 148.26 \space g \space Ca_3N_2 }{1 \space mole \space Ca_3N_2 } = 69.8 \space g \space Ca_3N_2 $$
Calculation of percent yield: $$\frac{ 32.4 \space g \space Ca_3N_2 }{ 69.8 \space g \space Ca_3N_2 } \times 100\% = 46.4 \%$$
