Question

What is the percent yield of NH3, if the following reaction is performed using 77 g CaCN, and produces 27.1 g NH,? CaCN2+3H2O +CaCO3+2NH3 Select the correct answer below. O 76,4% O 82.9% O 89.45

Answer 1

Molar mass of CaCN2,

MM = 1*MM(Ca) + 1*MM(C) + 2*MM(N)

= 1*40.08 + 1*12.01 + 2*14.01

= 80.11 g/mol

mass of CaCN2 = 77 g

mol of CaCN2 = (mass)/(molar mass)

= 77/80.11

= 0.9612 mol

According to balanced equation

mol of NH3 formed = (2/1)* moles of CaCN2

= (2/1)*0.9612

= 1.922 mol

Molar mass of NH3,

MM = 1*MM(N) + 3*MM(H)

= 1*14.01 + 3*1.008

= 17.034 g/mol

mass of NH3 = number of mol * molar mass

= 1.922*17.03

= 32.75 g

% yield = actual mass*100/theoretical mass

= 27.1*100/32.75

= 82.9 %

Answer: 82.9 %