Question

Calcium cyanamide reacts with water to form calcium carbonate and ammonia

CaCN2(s) + 3H2O(l)--------caCO3(s) .+ 2NH3(g)

Part a- how many grams of water are needed to react with 72.0g CaCN2?

Part b- how many grams of NH3 are produced from 5.19 g CaCN2?

Part c- how many grams of CaCO3 form if 151g water react?

Answer 1

Part a

1 mol CaCN2 require = 3 mol of H2O

80 g CaCN2 require = 3 x 18 = 54 g of H2O

72 g CaCN2 require = 54 x 72/80 = 48.6 g of H2O

Part b

1 mol CaCN2 require to produce = 2 mol of NH3

80 g CaCN2 require to produce = 2 x 17 = 34 g of NH3

5.19 g CaCN2 require to produce = 34 x 5.19/80 = 2.206 g of NH3

Part c

3 mol H2O require to produce = 1 mol of CaCO3

54 g of H2O require to produce = 100 g of CaCO3

151 g of H2O require to produce = 100 x 151/54 = 279.63 = 280 g of CaCO3