Question

1.46 g H2 is allowed to react with 10.1 g N2, producing 2.65 g NH3.

1. What is the theoretical yield in grams for this reaction under the given conditions?

Express your answer to three significant figures and include the appropriate units.

2. What is the percent yield for this reaction under the given conditions?

Express your answer to three significant figures and include the appropriate units.

Answer 1

1)

Molar mass of H2 = 2.016 g/mol

mass(H2)= 1.46 g

use:

number of mol of H2,

n = mass of H2/molar mass of H2

=(1.46 g)/(2.016 g/mol)

= 0.7242 mol

Molar mass of N2 = 28.02 g/mol

mass(N2)= 10.1 g

use:

number of mol of N2,

n = mass of N2/molar mass of N2

=(10.1 g)/(28.02 g/mol)

= 0.3605 mol

Balanced chemical equation is:

3 H2 + N2 ---> 2 NH3

3 mol of H2 reacts with 1 mol of N2

for 0.7242 mol of H2, 0.2414 mol of N2 is required

But we have 0.3605 mol of N2

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)*0.7242

= 0.4828 mol

use:

mass of NH3 = number of mol * molar mass

= 0.4828*17.03

= 8.224 g

Answer: 8.22 g

2)

% yield = actual mass*100/theoretical mass

= 2.65*100/8.224

= 32.22%

Answer: 32.2 %