Question

One method to produce nitrogen in the lab is to react ammonia with copper (II) oxide: NH3(
g. + CuO(s) Cu(s) + H2O(l) + N2(
g. After using 40.0 grams of NH3, 15.5 grams of N2 are produced. What is the percent yield of nitrogen in the reaction?

Answer 1

40g(1mol NH3/17.04g per mol), N2's molar mass=28.02g
3CuO+2NH3=>3Cu+N2+3H2O
(2.3473mol NH3)(1 mol N2/2 mol NH3)(28.02g/1mol)=32.8873g N2
15.5/32.8873=47.1306% yield

Answer 2

Answer:

47.05% is the percent yield of nitrogen in the reaction.

Explanation:

$2NH_3(g)+3CuO(s)\rightarrow 3Cu(s)+3H_2O(l) +N_2(g$

Theoretical yield of nitrogen gas = x

Moles of ammonia = $\frac{40.0 g}{17 g/mol}=2.3529 mol$

According to reaction,2 moles of ammonia gives 1 mol of nitrogen gas.

Then 2.3529 mol of ammonia will give:

$\frac{1}{2}\times 2.3529 mol=1.1764 mol$ of nitrogen gas

Mass of 1.1764 moles of nitrogen gas,x = 1.1764 mol × 28 g/mol=32.94 g

Experiential yield of nitrogen gas = 15.5 g

Percentage yield:

$\% yield=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$\% yield=\frac{15.5 g}{32.94 g}\times 100=47.05\%$

47.05% is the percent yield of nitrogen in the reaction.