Question

If 125.5g of Ca₃N₂ were produced from 29.0g of N₂ and an excess of Ca according to the reaction

3Ca + N_{2 ------------}Ca₃N₂

What was the percent yield of Ca₃N₂?

Answer 1

Molar mass of N2 = 28.02 g/mol

mass of N2 = 29 g

mol of N2 = (mass)/(molar mass)

= 29/28.02

= 1.035 mol

Balanced chemical equation is:

3 Ca + N2 —> Ca3N2

According to balanced equation

mol of Ca3N2 formed = moles of N2

= 1.035 mol

Molar mass of Ca3N2,

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

= 3*40.08 + 2*14.01

= 148.26 g/mol

mass of Ca3N2 = number of mol * molar mass

= 1.035*1.483*10^2

= 1.534*10^2 g

% yield = actual mass*100/theoretical mass

= 1.255*10^2*100/1.534*10^2

= 81.79 %

Answer: 81.79 %