Question

1. When 66.0 g of calcium is reacted with nitrogen gas, 28.1 g of calcium nitride is produced.

Part A

What is the percent yield of calcium nitride for this reaction?
3Ca(s)+N2(g)→Ca3N2(s)

Express your answer with the appropriate units.

2. In Part A, we saw that the theoretical yield of aluminum oxide is 1.70 mol . Calculate the percent yield if the actual yield of aluminum oxide is 1.14 mol .

Express your answer with the appropriate units.

Answer 1

(1) Given reaction is

3Ca(s) + N2(g) -----> Ca3N2(g)

Given mass of Ca=66 g and molar mass of Ca=40.078 g/mol.

Moles of Ca=mass/molar mass=66g/40.078 g/mol=1.646 mol.

Moles of Ca3N2 formed=(1/3) mol Can=1.646 mol/3=0.5489mol.

Molar mass of Ca3N2=148.25 g/mol.

Therefore theoretical mass of Ca3N2=moles x molar mass=0.5489 mol x 148.25 g/mol=81.379 g.

Given experimental mass of Ca3N2=28.1 g.

Then % yield=(experimental/theoretical) x 100

% yield=(28.1 g/81.379 g)x100=34.53 %.

(2) Given theoretical yield of aluminum oxide=1.70 mol

The experimental yield of aluminum oxide=1.14 mol.

Therefore % yield=(experimental/theoretical)x100=(1.14 mol/1.70 mol)x100=67.06 %.

Please let me know if you have any doubt. Thanks.