For the balanced equation shown below, if the reaction of 77.0 grams of CaCN2 produces 26.1 grams of NH3, what is the percent yield? CaCN2+3H2O=>CaCO3+2NH3

%yield = 86.93%

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%