Answer
287 g of $CaCO_3$
Work Step by Step
1. Calculate the molar mass of $H_2O$:
$H: 1.008g * 2= 2.016g $
$O: 16.00g$
2.016g + 16.00g = 18.02g
$ \frac{1 mole (H_2O)}{ 18.02g (H_2O)}$ and $ \frac{ 18.02g (H_2O)}{1 mole (H_2O)}$
2. The balanced reaction is:
$CaCN_2 + 3H_2O --\gt CaCO_3 + 2NH_3$
According to the coefficients, the ratio of $H_2O$ to $CaCO_3$ is 3 to 1:
$ \frac{ 1 mole(CaCO_3)}{ 3 moles (H_2O)}$ and $ \frac{ 3 moles (H_2O)}{ 1 mole(CaCO_3)}$
3. Calculate the molar mass for $CaCO_3$:
$Ca: 40.08g $
$C: 12.01g $
$O: 16.00g * 3= 48.00g $
40.08g + 12.01g + 48.00g = 100.09g
$ \frac{1 mole (CaCO_3)}{ 100.09g (CaCO_3)}$ and $ \frac{ 100.09g (CaCO_3)}{1 mole (CaCO_3)}$
4. Use the conversion factors to find the mass of $CaCO_3$
$155g(H_2O) \times \frac{1 mole(H_2O)}{ 18.02g( H_2O)} \times \frac{ 1 mole(CaCO_3)}{ 3 moles (H_2O)} \times \frac{ 100.09 g (CaCO_3)}{ 1 mole (CaCO_3)} = 287g (CaCO_3)$
