(45 g CaCN2) / (80.1026 g CaCN2/mol) = 0.5618 mol CaCN2
(45 g H2O) / (18.0153 g H2O/mol) = 2.498 mol H2O
0.5618 mole of CaCN2 would react completely with 0.5618 x (3/1) = 1.685 mol H2O, but there is more H2O present than that, so H2O is in excess and CaCN2 is the limiting reactant.
(0.5618 mol CaCN2) x (1/1) x (100.0875 g CaCO3/mol) = 56 g CaCO3
Answers & Comments
Verified answer
CaCN2 + 3 H2O → CaCO3 + 2 NH3
(45 g CaCN2) / (80.1026 g CaCN2/mol) = 0.5618 mol CaCN2
(45 g H2O) / (18.0153 g H2O/mol) = 2.498 mol H2O
0.5618 mole of CaCN2 would react completely with 0.5618 x (3/1) = 1.685 mol H2O, but there is more H2O present than that, so H2O is in excess and CaCN2 is the limiting reactant.
(0.5618 mol CaCN2) x (1/1) x (100.0875 g CaCO3/mol) = 56 g CaCO3