%yield = 86.93%
Further explanation
Given
Reaction
CaCN₂+3H₂O=>CaCO₃+2NH₃
Required
The percent yield
Solution
mol CaCN₂(MW= 92 g/mol) :
= mass : MW
= 77 g : 92 g/mol
= 0.834
From equation, mol NH₃ :
= 2/1 x mol CaCN₂
= 2/1 x 0.834
= 1.668
Mass NH₃(theoretical):
= 1.668 mol x 18 g/mol
= 30.024 g
% yield = (actual/theoretical) x 100%
%yield = (26.1/30.024) x 100%
%yield = 86.93%