Answer
50.6 g of $H_2O$
Work Step by Step
1. Calculate the molar mass of $CaCN_2$:
$Ca: 40.08g $
$C: 12.01g $
$N: 14.01g * 2= 28.02g $
40.08g + 12.01g + 28.02g = 80.11g
$ \frac{1 mole (CaCN_2)}{ 80.11g (CaCN_2)}$ and $ \frac{ 80.11g (CaCN_2)}{1 mole (CaCN_2)}$
2. The balanced reaction is:
$CaCN_2 + 3H_2O --\gt CaCO_3 + 2NH_3$
According to the coefficients, the ratio of $CaCN_2$ to $H_2O$ is 1 to 3:
$ \frac{ 3 moles(H_2O)}{ 1 mole (CaCN_2)}$ and $ \frac{ 1 mole (CaCN_2)}{ 3 moles(H_2O)}$
3. Calculate the molar mass for $H_2O$:
$H: 1.008g * 2= 2.016g $
$O: 16.00g$
2.016g + 16.00g = 18.02g
$ \frac{1 mole (H_2O)}{ 18.016g (H_2O)}$ and $ \frac{ 18.02g (H_2O)}{1 mole (H_2O)}$
4. Use the conversion factors to find the mass of $H_2O$
$75.0g(CaCN_2) \times \frac{1 mole(CaCN_2)}{ 80.11g( CaCN_2)} \times \frac{ 3 moles(H_2O)}{ 1 mole (CaCN_2)} \times \frac{ 18.016 g (H_2O)}{ 1 mole (H_2O)} = 50.6g (H_2O)$
