Question

1.16 g H2 is allowed to react with 9.82 g N2, producing 2.06 g NH3.

A) What is the theoretical yield in grams for this reaction under the given conditions?

B) What is the percent yield for this reaction under the given conditions?

Answer 1

1)

Molar mass of N2 = 28.02 g/mol

mass(N2)= 9.82 g

use:

number of mol of N2,

n = mass of N2/molar mass of N2

=(9.82 g)/(28.02 g/mol)

= 0.3505 mol

Molar mass of H2 = 2.016 g/mol

mass(H2)= 1.16 g

use:

number of mol of H2,

n = mass of H2/molar mass of H2

=(1.16 g)/(2.016 g/mol)

= 0.5754 mol

Balanced chemical equation is:

N2 + 3 H2 ---> 2 NH3

1 mol of N2 reacts with 3 mol of H2

for 0.3505 mol of N2, 1.051 mol of H2 is required

But we have 0.5754 mol of H2

so, H2 is limiting reagent

we will use H2 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 = (2/3)* moles of H2

= (2/3)*0.5754

= 0.3836 mol

use:

mass of NH3 = number of mol * molar mass

= 0.3836*17.03

= 6.534 g

Answer: 6.53 g

2)

% yield = actual mass*100/theoretical mass

= 2.06*100/6.534

= 31.53%

Answer: 31.5 %