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)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.     1.98 gg H2H2 is allowed to react with 9.76 gg N2N2, producing 1.95 gg NH3NH3. 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.   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.

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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)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.

 
 
1.98 gg H2H2 is allowed to react with 9.76 gg N2N2, producing 1.95 gg NH3NH3.

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.
 

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.
Expert Solution
Step 1

When ammonia react with hydrogen then it will form ammonia so first write balanced reaction.

N2+3H22NH3

Molar mass of nitrogen atom, hydrogen atom and ammonia is 14 g/mol, 1 g/mol and 17 g/mol. From above reaction, 28 g of nitrogen react with 6 g of hydrogen. First calculate limiting reagent, for this calculate how much mass of nitrogen is required for 1.98 g of hydrogen, if calculated mass of nitrogen required more than the 9.76 g, which is present initially, then nitrogen is limiting reagent. If it comes less than 9.76 g then hydrogen is limiting reagent. 

For calculation of mass of nitrogen required, first divide molar mass of nitrogen to molar mass of hydrogen then multiplied it with mass of hydrogen.

 Mass of N2 Required=286×1.98 g=9.24 g

Mass of nitrogen required is 9.24 g for 1.98 g of hydrogen. Initially 9.76 g of nitrogen is present which is greater than mass of nitrogen required hence hydrogen is limiting reagent. 

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