The Haber-Bosch process is a very important industrial process. In the Haber-Bos

Question

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

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.

1.19 g H2 is allowed to react with 9.90 g N2, producing 2.44 g NH3.

Part A

What is the theoretical yield for this reaction under the given conditions?

Part B

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

Explanation / Answer

   3H2(g)+N2(g) ---> 2NH3(g)

  
No of moles of H2 =wt/mwt = 1.19/2 = 0.595 mole

No of moles of N2 = 9.9/28 = 0.3535 mole

limiting reagent is H2

No moles of NH3 produced = 0.595*2/3 = 0.397 mole

mass of NH3 produced = 0.397*17 = 6.749 grams

part A

theoretical yield = 6.749 grams NH3

practical yield = 2.44 grams NH3.

part B

percent yield = practical yield/theoretical yield*100

= 2.44/6.749*100

   = 36.15%

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.