CaCN2 + 3 H2O → CaCO3 + 2 NH3 QUESTION IN DESCRIPTION TOO LONG

Question

Given the reaction:

CaCN2 + 3 H2O → CaCO3 + 2 NH3

What is the percent yield if 15.5 g of NH3 is produced upon reaction of 65.5 g of CaCN2 and 451 g of water? The molar mass of CaCN2 is 80.11 g/mol and the molar mass of CaCO3 is 100.09 g/mol.

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Answer 1

CaCN2 + 3 H2O → CaCO3 + 2 NH3

Limiting reactant:

65.5 g CaCN2 x 1 mol/80.11 g = 0.8176 moles CaCN2

451 g H2O x 1 mol/18 g = 20.06 moles H2O

Since there is more that 3x as much H2O as CaCN2, the CaCN2 is limiting (see mol ratio in balanced eq.)

Once we know the limiting reactant, we can find the theoretical yield of NH3:

0.8176 moles CaCN2 x 2 mol NH3/mol CaCN2 x 17 g NH3/mol NH3 = 27.8 g NH3 = theoretical yield

% yield NH3 = actual yield/theoretical yield (x100%) = 15.5 g/27.8 g (x100%) = 55.8% yield