(Solved) Percent Yield

joel189

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11 years ago

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) Rightwards Arrow

2NH3 (g)

1.63g H2 is allowed to react with 10.5g N2, producing 1.84g NH3.

Part A: What is the theoretical yield for this reaction under the given conditions? (Express your answer numerically in grams.)

Part B: What is the percent yield for this reaction under the given conditions? (Express the percentage numerically.)

Source Pearson Mastering Chemistry

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etaylor6

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11 years ago

Sample

1.93 g H_2 is allowed to react with 9.81 g N_2, producing 1.55 g NH_3.

3H_2(g) + N_2(g) ----> 2NH_3(g)

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

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

Solution

Your requirements are

3 moles H2 to 1 mole N2 to yield 2 moles of NH3

1.93 g H2 = 0.965 moles H2
9.81 g N2 = 0.350 moles N2
1.55 g NH3 = 0.0912 moles NH3

The ratio of H2 to N2 is 2.76, but you need 3, so there is not enough H2; H2 is the limiting reagent and will determine the amount of NH3 produced. Each mole of H2 yields 2/3 moles of NH3, so you should get 0.965*(2/3) = 0.643 moles (10.9 g) of NH3, and that is the theoretical yield. The actual yield is 0.0912 moles, so the percent yield is 0.0912/0.643 * 100 = 14.2%

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etaylor6

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11 years ago

Sample

1.69g of H2(subscript) is allowed to react with 9.64g N2(subscript), producing 1.63g NH3(subscript).

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

b. what is the percent yield for this reation under the given conditions?

Solution

The reaction is 3H2 + N2 ----> 2NH3
=> 3*2 g Hydrogen + 1*28 g of Nitrogen = 2*17 g of Ammonia
=> 6 g of Hydrogen + 28 g of Nitrogen = 34 g of Ammonia

=> 6 g of Hydrogen need Nitrogen = 28 g
=> 1 g of Hydrogen need Nitrogen = 28/6 g

Now given data
Hydrogen = 1.69
1.69 g of Hydrogen need Nitrogen = 28/6*1.69 = 7.887 g
Available Nitrogen = 9.64 g

Therefore Hydrogen is the limiting reactant and will consume itself before Nitrogen is consumed and the calculations will be based on Hydrogen.

Theoretical Yield:
6 g Hydrogen produces Ammonia = 34 g
=> 1 g Hydrogen produces Ammonia = 34/6 g
=> 1.69 g Hydrogen produces Ammonia = 34/6*1.69 = 9.577 g
This is the Theoretical Yield of this reaction

Percentage Yield = Actually Produced/Theoretically produced * 100
Percentage Yield = 1.63/9.577*100 = 17.02%

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joel189 Author

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Percent Yield

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