N2(g) + H2(g) → NH3(g) If you started with 2.5 g of N2 and 4.5 g of H2, how many grams of NH3 will you make?

Answer 1

Hello, Addy,

We need to have a balanced equation.

N₂ + H₂ = NH₃

We can see from the NH₃ that we need 2 of them to account for the two nitrogens from N₂, so add a coefficient of 2. This creates a need for 6 hydrogens, so place a 3 for the H₂, and the equation now looks balanced.

N₂ + 3H₂ = 2NH₃

Calculate moles from the masses of both the N₂ and H₂ . Do this by dividing the masses by the molar mass of each reactant.

N₂ : 2.5g/(28g/mole) = 0.089 moles

H₂ : 4.5g/(2g/mole) = 2.25 moles

The balanced equation says, for H₂ we need three times the moles of N₂ . That's 0.268 moles of hydrogen that we'll need to react with all of the nitrogen. We have far more than that of H₂ , so that means that the N₂ is the limiting reagent. We will assume of of it is used in making the ammonia (NH₃).

Multiply the moles of N₂ by 3, since the equation says we'll get 3 moles of ammonia for every mole of nitrogen, and it is moles of ammonia we want. That yields 0.178 moles of NH₃. Multiply that by ammonia's molar mass of 17 to find 3.04 grams of ammonia. Trim that to 3.0 grams for two sig figs.

As a side note, we are left with 1.98 moles of unreacted H₂. That's 4 grams of hydrogen. Added to the 3 grams of ammonia, that means we have 7 grams total of products and unreacted reactants. We started with a total of (2.5 + 3.5) 7 grams of reactants, so the conservation of matter law is demonstrated here.

I hope this helps,

Bob