How many grams of CaCN2 are required in order to produce 4.97 grams of ammonia (NH3)? Balanced Equation: 1CaCN2 (s) + 3 H2O (l) ---------> 1CaCO3 (s) + 2NH3 (g)

Answer 1

CaCN2 (s) + 3 H2O (l) ---------> CaCO3 (s) + 2NH3 (g)

4.97 g NH3 x 1 mol NH3/17 g = 0.292 moles NH3

0.292 mol NH3 x 1 mol CaCN2/2 mol NH3 = 0.146 mol CaCN2

grams CaCN2 = 0.146 mol x 80.1 g/mol = 11.7 g