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.
1.16 g H2 is allowed to react with 10.4 g N2, producing 2.93 g NH.
What is the theoretical yield in grams for this reaction under the given conditions?
To find the theoretical yield of ammonia, we first need to determine the limiting reactant in the reaction. This can be done by comparing the amount of product produced by each reactant.
First, we need to calculate the moles of each reactant:
Moles of H2 = 1.16 g / 2.02 g/mol = 0.574 moles
Moles of N2 = 10.4 g / 28.02 g/mol = 0.371 moles
Next, we need to determine which reactant limits the amount of product produced. This can be done by calculating the moles of ammonia that can be produced from each reactant:
From H2: 0.574 moles H2 x (2 moles NH3 / 3 moles H2) = 0.383 moles NH3
From N2: 0.371 moles N2 x (2 moles NH3 / 1 mole N2) = 0.742 moles NH3
Since the moles of ammonia produced from H2 is less than the moles produced from N2, H2 is the limiting reactant in this reaction.
Now, we can calculate the theoretical yield of ammonia using the limiting reactant:
Theoretical yield of NH3 = 0.383 moles NH3 x 17.03 g/mol = 6.517 g
Therefore, the theoretical yield of ammonia under the given conditions is 6.517 grams.