Question

When  97.4 g of calcium and 44.7 g of nitrogen gas undergo a reaction that has a  89.4% yield, what mass of calcium nitride forms?

______ g calcium nitride

Answer 1

Molar mass of Ca = 40.08 g/mol

mass(Ca)= 97.4 g

number of mol of Ca,

n = mass of Ca/molar mass of Ca

=(97.4 g)/(40.08 g/mol)

= 2.43 mol

Molar mass of N2 = 28.02 g/mol

mass(N2)= 44.7 g

number of mol of N2,

n = mass of N2/molar mass of N2

=(44.7 g)/(28.02 g/mol)

= 1.595 mol

Balanced chemical equation is:

3 Ca + N2 ---> Ca3N2

3 mol of Ca reacts with 1 mol of N2

for 2.43014 mol of Ca, 0.810047 mol of N2 is required

But we have 1.595289 mol of N2

so, Ca is limiting reagent

we will use Ca in further calculation

Molar mass of Ca3N2,

MM = 3*MM(Ca) + 2*MM(N)

= 3*40.08 + 2*14.01

= 148.26 g/mol

According to balanced equation

mol of Ca3N2 formed = (1/3)* moles of Ca

= (1/3)*2.43014

= 0.810047 mol

mass of Ca3N2 = number of mol * molar mass

= 0.81*1.483*10^2

= 120.1 g

% yield = actual mass*100/theoretical mass

89.4= actual mass*100/120.1

actual mass=107 g

Answer: 107 g