Question

1.28 g H2 is allowed to react with 9.54 g N2, producing 1.31 g NH3.

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

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

Answer 1

A)

Molar mass of N2 = 28.02 g/mol

mass of N2 = 9.54 g

we have below equation to be used:

number of mol of N2,

n = mass of N2/molar mass of N2

=(9.54 g)/(28.02 g/mol)

= 0.3405 mol

Molar mass of H2 = 2.016 g/mol

mass of H2 = 1.28 g

we have below equation to be used:

number of mol of H2,

n = mass of H2/molar mass of H2

=(1.28 g)/(2.016 g/mol)

= 0.6349 mol

we have the Balanced chemical equation as:

N2 + 3 H2 ---> 2 NH3 +

1 mol of N2 reacts with 3 mol of H2

for 0.3405 mol of N2, 1.021 mol of H2 is required

But we have 0.6349 mol of H2

so, H2 is limiting reagent

we will use H2 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 3 mol of H2 reacts, 2 mol of NH3 is formed

mol of NH3 formed = (2/3)* moles of H2

= (2/3)*0.6349

= 0.4233 mol

we have below equation to be used:

mass of NH3 = number of mol * molar mass

= 0.4233*17.03

= 7.21 g

Answer: 7.21 g

B)

% yield = actual mass*100/theoretical mass

= 1.31*100/7.21

= 18.2 %

Answer: 18.2 %