Nitrogen and hydrogen reacted to form 4.00 g of NH3 with 35.5 % yield. If 1.05 g of nitrogen remained unreacted, how much nitrogen must have been present at the start?

Answer 1

Begin with a correctly balanced equation for the reaction taking place:

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

Let's determine moles of NH₃ produced:

4.00 g NH₃ x 1 mol NH₃/17.0 g = 0.235 moles NH₃

Since this represents only 35.5%, to find what 100% would be, we simply convert to 100%:

0.235 moles / 0.355 = 0.662 moles NH₃ = THEORETICAL YIELD

We are told that 1.05 g of N₂ remained unreacted, so that means it was in excess and H₂ was limiting.

From the balanced equation, we see that 3 mol H₂ produces 2 mol NH₃ so we can use this to find moles H₂ originally present:

0.662 mol NH₃ x 3 mol H₂/2 mol NH₃ = 0.993 moles H₂ present

Finding moles N₂ that reacted, we have:

0.993 mol H₂ x 1 mol N₂/3 mol H₂ = 0.331 mol N₂ reacted and this is 0.331 mol x 28 g/mol = 9.27g N₂

The amount of N₂ left is 1.05 g, so the amount of N₂ originally present would be 9.27 g + 1.05 g = 10.3 g N₂