1.48 g H2 is allowed to react with 10.0 g N2, producing 1.83 g NH3.

Question

1.48 g H2 is allowed to react with 10.0 g N2, producing 1.83 g NH3.

 

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.

Answer 1

Hello Areej

 

The balanced chemical reaction is:

 

3H₂ (g) + N₂ (g) → 2NH₃ (g)

 

PART A

 

As the other tutor Angie mentioned, you have to first determine whether H₂ or N₂ is the limiting reagent as the theoretical yield of NH₃ is dependent upon the # of moles of the limiting reagent.

 

You have to use stoichiometry to perform these calculations.  To convert grams to moles, we need the molar mass of H₂ (2 g/mol) and N₂ (28 g/mol).  From the reaction, you can see that 3 moles of H₂ reacts to form 2 moles of NH₃, and 1 mole of N₂ reacts to form 2 moles of NH₃.  The process is as follows:

 

g H2 → mol H2 → mol NH3

 

g N2 → mol N2 → mol NH₃

 

1.48 g H₂ x (1 mol H₂ / 2 g H₂) x (2 mol NH₃ / 3 mol H₂) = 0.493 mol NH₃

 

10.0 g N₂ x (1 mol N₂ / 28 g N₂) x (2 mol NH₃ / 1 mol N₂) = 0.714 mol NH₃

 

From above, you can see that H2 produces less NH3, therefore, it is the limiting reagent.  The theoretical yield of NH3 is:

 

0.493 mol NH₃ x (17 g NH₃ / 1 mol NH₃) = 8.38 g NH₃

 

PART B

 

%-Yield = (actual yield / theoretical yield) x 100%

  

            = 1.83 / 8.38 x 100% = 21.8%

Answer 2

Expanding on the answer given by Angie B, look at the balanced equation.

3H₂ + N₂ ==> 2NH₃  (Haber reaction)

moles H₂ present = 1.48 g H₂ x 1 mole/2 g = 0.74 moles H₂

moles  N₂ present = 10.0 g N₂ x 1 mole/28 g = 0.357 moles N₂

Limiting reactant is H₂ since it will run out first based on the ratio of 3 moles H₂ to 1 mole N₂

(A)  Theoretical yield = 0.74 moles H₂ x 2 moles NH₃/3 moles H₂ = 0.493 moles NH₃ x 17 g/mole = 8.39 g NH₃

(B)  Percent yield = actual yield/theoretical yield (x100%) = 1.83 g/8.39 g (x100%) = 21.8% yield

Answer 3

First, you should decide which of the reactants is limiting. The easiest way to do that is to convert one of the reactants to the other and see which will run out first.