Question

CH 06 HW t Percent Yield 3 of 7 Constants Periodic Table The Haber-Bosch process is a very important industrial process. In the Haber- Bosch process, hydrogen gas reacts with nitrogen gas to produce ammonia according to the equation 1.79 g H2 is allowed to react with 10.2 g N2, producing 1.21 g NH3 3H2 (g)N2 (g)2NH3(g) The ammonia produced in the Haber-Bosch process has a wide range of uses, from fertilizer to pharmaceuticals. However, the production of ammonia is difficult resulting in lower yields than those predicted from the chemical equation Part A 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. View Available Hint(s) HA Value Units Submit Part B What is the percent yield for this reaction under the given conditions? Express your answer to three significant figures and include the appropriate units. View Available Hint(s) НА Value Units Submit Provide Feedback Next

i am having trouble with this question if anyone could help that would be great!

Answer 1

A)

Molar mass of H2 = 2.016 g/mol

mass(H2)= 1.79 g

use:

number of mol of H2,

n = mass of H2/molar mass of H2

=(1.79 g)/(2.016 g/mol)

= 0.8879 mol

Molar mass of N2 = 28.02 g/mol

mass(N2)= 10.2 g

use:

number of mol of N2,

n = mass of N2/molar mass of N2

=(10.2 g)/(28.02 g/mol)

= 0.364 mol

Balanced chemical equation is:

3 H2 + N2 ---> 2 NH3 +

3 mol of H2 reacts with 1 mol of N2

for 0.8879 mol of H2, 0.296 mol of N2 is required

But we have 0.364 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)*0.8879

= 0.5919 mol

use:

mass of NH3 = number of mol * molar mass

= 0.5919*17.03

= 10.08 g

Answer: 10.1 g

B)

% yield = actual mass*100/theoretical mass

= 1.21*100/10.08

= 12.0 %

Answer: 12.0 %