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.

1.43 g H2 is allowed to react with 9.70 g N2, producing 2.31 g NH3.

Part A: What is the theoretical yield in grams for this reaction under the given conditions?

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

Answer 1

balanced reaction given below using it we can find out theoretical yield

N2 + 3H2 -->2 NH3

mass of N2 = 9.70 g, we have molar mass of N2 =28.0134 g/mol

hence moles of N2 present = given mass/molar mass= 9.70g/28.0134 g/mol =0.3466 mols

mass of H2 = 1.43 g, molar mass of H2 =2.016 g/mol

moles of H2 present = 1.43 g/2.016 g/mol = 0.70933 moles

mole ratio N2 :H2 = 1:3

Hence moles of H2 needed is 3 times that of N2. But fewer moles of H2 is present .SO limiting reactant is

H2. moles of NH3 formed depends on moles of H2.

H2:NH3 = 3:2

moles of NH3 formed = [2/3]*0.70933 moles 0.47289 mols

molar mass of NH3 = 17.03 g/moles

hence mass of product formed = #mols of NH3*molar mass =0.70933 moles*17.03 g/mol

=8.0533 g

thus theoretical yield of NH3 =8.0533 g

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Actual yield of NH3 = 2.31 g

% yield = actual yield/ theoretical yield *100 = 2.31 g/8.0533 g*100 = 28.68 %

round to 3 significant figures

% yield = 28.7 %

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:) thanks

Answer 2

What is the theoretical yield in grams for this reaction under the given conditions? 5.97 g

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

20.9%