Question

Given N₂ + 3 H₂ → 2 NH₃

If 5.55 g of each reactant are used, what is the theoretical mass, in grams, of ammonia that will be produced?

Also, what is the percent yield for this reaction if 5.55 g of ammonia are obtained experimentally?

Answer 1

1)

Molar mass of N2 = 28.02 g/mol

mass(N2)= 5.55 g

use:

number of mol of N2,

n = mass of N2/molar mass of N2

=(5.55 g)/(28.02 g/mol)

= 0.1981 mol

Molar mass of H2 = 2.016 g/mol

mass(H2)= 5.55 g

use:

number of mol of H2,

n = mass of H2/molar mass of H2

=(5.55 g)/(2.016 g/mol)

= 2.753 mol

Balanced chemical equation is:

N2 + 3 H2 ---> 2 NH3

1 mol of N2 reacts with 3 mol of H2

for 0.1981 mol of N2, 0.5942 mol of H2 is required

But we have 2.753 mol of H2

so, N2 is limiting reagent

we will use N2 in further calculation

Molar mass of NH3,

MM = 1*MM(N) + 3*MM(H)

= 1*14.01 + 3*1.008

= 17.034 g/mol

According to balanced equation

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

= (2/1)*0.1981

= 0.3961 mol

use:

mass of NH3 = number of mol * molar mass

= 0.3961*17.03

= 6.748 g

Answer: 6.75 g

2)

% yield = actual mass*100/theoretical mass

= 5.55*100/6.748

= 82.25%

Answer: 82.25 %