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 yi
eld?
CaCN2+3H2O=>CaCO3+2NH3
1 answer:
%yield = 86.93%
<h3>Further explanation</h3>
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%
<em>%yield = 86.93%</em>
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