t Percent Yield 5015 > The Haber-Bosch process is a very important industrial pr

Question

t Percent Yield 5015 > 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 (8) + N2(8) 2NHs() 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 1.04 g H2 is allowed to react with 9.69 gN2. producing 2.47 g NHs 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 Submit

Explanation / Answer

Ans :

Part A :

Number of moles of H2 = 1.04 / 2.016 = 0.516 mol

Number of moles of N2 = 9.69 / 28.0134 = 0.346 mol

one mole of N2 requires three mol of H2

So 0.346 mol of N2 will require : 3 x 0.346 = 1.04 mol of H2

So H2 is the limiting reagent

Now 3 moles of H2 make 2 moles of NH3

So 0.516 mol of H2 will produce : (0.516 x 2) / 3 = 0.344 mol of NH3

Theoretical yield = 0.344 x molar mass NH3

= 0.344 x 17.03

= 5.86 g

Part B :

Percent yield = ( actual yield / theoretical yield) x 100

= (2.47 / 5.86) x 100

= 42.2 %

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