The ammonia produced in the Haber-Bosch process has a wide rangeof uses, from fe

Question

The ammonia produced in the Haber-Bosch process has a wide rangeof uses, from fertilizer to pharmaceuticals. However, theproduction of ammonia is difficult, resulting in lower yields thanthose predicted from the chemical equation.

The Haber - Bosch process is a very important industrial process. Inthe Haber - Bosch process, hydrogen gas reacts with nitrogen gas toproduce ammonia according to the equation NH_3 What is the theoretical yield for this reaction under the givenconditions? What is the percent yield for this reaction under the givenconditions?g N_2, producing 1.41 g H_2 is allowed to react with10.4 g 1.45 3H_2(g)+N_2(g)rightarrow2NH_3(g) The ammonia produced in the Haber - Bosch process has a wide rangeof uses, from fertilizer to pharmaceuticals. However, theproduction of ammonia is difficult, resulting in lower yields thanthose predicted from the chemical equation.

Explanation / Answer

Molar mass of NH3 = 14 + 3 * 1 = 17 g Molar mass of N2 = 2 * 14 = 28 g Molar mass of H2 = 2 * 1 = 2 g 3* 2 g of H2 reacts with 28 g of N2 1.45 g of H2 reacts with X g of N2 X = ( 1.45 * 28 ) / 6     = 6.767 g of N2 So 10.4 - 6.767 g of N2 is left unreacted 6 g of H2 produces 2 * 17 g of NH3 1.45  g of H2 produces Yg of NH3 Y = ( 1.45 * 2 * 17 ) / 6     = 8.216 g ofNH3              ------- theoretical yield % yield = ( actual yield / theoretical yield ) *100            = ( 1.41 g / 8.216 ) *100            = 17.16 % 1.45 g of H2 reacts with X g of N2 X = ( 1.45 * 28 ) / 6     = 6.767 g of N2 So 10.4 - 6.767 g of N2 is left unreacted 6 g of H2 produces 2 * 17 g of NH3 1.45  g of H2 produces Yg of NH3 Y = ( 1.45 * 2 * 17 ) / 6     = 8.216 g ofNH3              ------- theoretical yield % yield = ( actual yield / theoretical yield ) *100            = ( 1.41 g / 8.216 ) *100            = 17.16 % 6 g of H2 produces 2 * 17 g of NH3 1.45  g of H2 produces Yg of NH3 Y = ( 1.45 * 2 * 17 ) / 6     = 8.216 g ofNH3              ------- theoretical yield % yield = ( actual yield / theoretical yield ) *100            = ( 1.41 g / 8.216 ) *100            = 17.16 %

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