Question

What is the theoretical yield of ammonia (in grams) if 17.15 grams of nitrogen gas and 10.95 grams of hydrogen gas are allowed to react?Based on your theoretical yield, what is the percent yield of ammonia if only 11.5 grams of ammonia is produced?

Answer 1

1)

Molar mass of N2 = 28.02 g/mol

mass of N2 = 17.15 g

we have below equation to be used:

number of mol of N2,

n = mass of N2/molar mass of N2

=(17.15 g)/(28.02 g/mol)

= 0.6121 mol

Molar mass of H2 = 2.016 g/mol

mass of H2 = 10.95 g

we have below equation to be used:

number of mol of H2,

n = mass of H2/molar mass of H2

=(10.95 g)/(2.016 g/mol)

= 5.432 mol

we have the Balanced chemical equation as:

N2 + 3 H2 ---> 2 NH3

1 mol of N2 reacts with 3 mol of H2

for 0.6121 mol of N2, 1.836 mol of H2 is required

But we have 5.432 mol of H2

so, N2 is limiting reagent

we will use N2 in further calculation

Molar mass of NH3 = 1*MM(N) + 3*MM(H)

= 1*14.01 + 3*1.008

= 17.034 g/mol

From balanced chemical reaction, we see that

when 1 mol of N2 reacts, 2 mol of NH3 is formed

mol of NH3 formed = (2/1)* moles of N2

= (2/1)*0.6121

= 1.224 mol

we have below equation to be used:

mass of NH3 = number of mol * molar mass

= 1.224*17.03

= 20.85 g

Answer: 20.85 g

2)

% yield = actual mass*100/theoretical mass

= 11.5*100/20.85

= 55.15%

Answer: 55.15%