Answer
a. 50.6 g of $H_2O$
b. 2.23 g of $NH_3$
c. 287 g of $CaCO_3$
Work Step by Step
a.
$$ \frac{ 1 \space mole \space CaCN_2 }{ 3 \space moles \space H_2O } \space and \space \frac{ 3 \space moles \space H_2O }{ 1 \space mole \space CaCN_2 }$$
$ CaCN_2 $ : ( 40.08 $\times$ 1 )+ ( 12.01 $\times$ 1 )+ ( 14.01 $\times$ 2 )= 80.11 g/mol
$$ \frac{1 \space mole \space CaCN_2 }{ 80.11 \space g \space CaCN_2 } \space and \space \frac{ 80.11 \space g \space CaCN_2 }{1 \space mole \space CaCN_2 }$$
$ H_2O $ : ( 1.008 $\times$ 2 )+ ( 16.00 $\times$ 1 )= 18.02 g/mol
$$ \frac{1 \space mole \space H_2O }{ 18.02 \space g \space H_2O } \space and \space \frac{ 18.02 \space g \space H_2O }{1 \space mole \space H_2O }$$
$$ 75.0 \space g \space CaCN_2 \times \frac{1 \space mole \space CaCN_2 }{ 80.11 \space g \space CaCN_2 } \times \frac{ 3 \space moles \space H_2O }{ 1 \space mole \space CaCN_2 } \times \frac{ 18.02 \space g \space H_2O }{1 \space mole \space H_2O } = 50.6 \space g \space H_2O $$
b. $$ \frac{ 1 \space mole \space CaCN_2 }{ 2 \space moles \space NH_3 } \space and \space \frac{ 2 \space moles \space NH_3 }{ 1 \space mole \space CaCN_2 }$$
$ CaCN_2 $ : ( 40.08 $\times$ 1 )+ ( 12.01 $\times$ 1 )+ ( 14.01 $\times$ 2 )= 80.11 g/mol
$$ \frac{1 \space mole \space CaCN_2 }{ 80.11 \space g \space CaCN_2 } \space and \space \frac{ 80.11 \space g \space CaCN_2 }{1 \space mole \space CaCN_2 }$$
$ NH_3 $ : ( 1.008 $\times$ 3 )+ ( 14.01 $\times$ 1 )= 17.03 g/mol
$$ \frac{1 \space mole \space NH_3 }{ 17.03 \space g \space NH_3 } \space and \space \frac{ 17.03 \space g \space NH_3 }{1 \space mole \space NH_3 }$$
$$ 5.24 \space g \space CaCN_2 \times \frac{1 \space mole \space CaCN_2 }{ 80.11 \space g \space CaCN_2 } \times \frac{ 2 \space moles \space NH_3 }{ 1 \space mole \space CaCN_2 } \times \frac{ 17.03 \space g \space NH_3 }{1 \space mole \space NH_3 } = 2.23 \space g \space NH_3 $$
c. $$ \frac{ 3 \space moles \space H_2O }{ 1 \space mole \space CaCO_3 } \space and \space \frac{ 1 \space mole \space CaCO_3 }{ 3 \space moles \space H_2O }$$
$ H_2O $ : ( 1.008 $\times$ 2 )+ ( 16.00 $\times$ 1 )= 18.02 g/mol
$$ \frac{1 \space mole \space H_2O }{ 18.02 \space g \space H_2O } \space and \space \frac{ 18.02 \space g \space H_2O }{1 \space mole \space H_2O }$$
$ CaCO_3 $ : ( 40.08 $\times$ 1 )+ ( 12.01 $\times$ 1 )+ ( 16.00 $\times$ 3 )= 100.09 g/mol
$$ \frac{1 \space mole \space CaCO_3 }{ 100.09 \space g \space CaCO_3 } \space and \space \frac{ 100.09 \space g \space CaCO_3 }{1 \space mole \space CaCO_3 }$$
$$ 155 \space g \space H_2O \times \frac{1 \space mole \space H_2O }{ 18.02 \space g \space H_2O } \times \frac{ 1 \space mole \space CaCO_3 }{ 3 \space moles \space H_2O } \times \frac{ 100.09 \space g \space CaCO_3 }{1 \space mole \space CaCO_3 } = 287 \space g \space CaCO_3 $$
