1.96 g H2 is allowed to react with 10.3 g N2, producing 1.61 g NH3.

Question

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

The first thing we must do is to write the correctly balanced equation:

3H₂ + N₂ ==> 2NH₃ ... balanced equation

Next, since we are given the mass of both reactants, we must determine which, if either, if present in limiting supply:

1.96 g H₂ x 1 mole H₂/2 g = 0.98 moles H₂ present

10.3 g N₂ x 1 mol N₂/28 g = 0.368 moles N₂ present

An easy way to find which is limiting is to divide the moles by the coefficient in the balanced equation...

For H₂, we have 0.98 mol/3 = 0.326

For N₂, we have 0.368/1 = 0.368

Since the value for H₂ is less, H₂ is LIMITING.

a) Theoretical yield of NH₃ = 0.98 moles H₂ x 2 mol NH₃/3 mol H₂ x 17 g NH₃/mol = 11.1 g NH₃

b) Percent yield = actual/theoretical (x100%) = 1.61 g/11.1 g (x100%) = 14.5% yield