A 30.6 g sample of nitrogen gas combines completely with 6.6 g of hydrogen gas to form ammonia. What is the mass of ammonia formed?

Question

A 30.6 g sample of nitrogen gas combines completely with 6.6 g of hydrogen gas to form ammonia. What is the mass of ammonia formed?

Answer 1

As a follow up to the answer given by Mr. d., and to provide some additional aid for you to solve this problem, I offer the following:

The balanced equation for the reaction taking place is...

N₂(g) + 3H₂(g) ==> 2NH₃(g)

moles N₂ present = 30.6 g N₂ x 1 mole N₂/28 g = 1.09 moles N₂ = 1.1 moles N₂

moles H₂ present = 6.6 g H₂ x 1 mole H₂/2.0 g = 3.3 moles H₂

NOTE that the mole ratio in the balanced equation is 3 moles H₂ for each 1 mole N₂ so in this case, neither is limiting because you have precisely 3 times as many moles of H₂ as you do N₂ (to 2 sig. figs. anyway).

To find the mass of NH₃ formed, use either reactant (since neither is limiting) and calculate mass of NH₃:

1.1 moles N₂ x 2 moles NH₃/mole N₂ x 17 g NH₃/mole NH₃ = grams NH₃

3.3 moles H₂ x 2 moles NH₃/3 moles H₂ x 17 g NH₃/mole NH₃ = grams NH₃

Answer 2

Hi Kendall T.,

You aren't going to like how roundabout this is, but bear with it!

Convert the given masses into moles of each gas. From the balanced equation for reaction, figure out if either reactant gas is present in excess; if so, use the limiting gas to carry through in to ammonia product, and then convert back into mass.

If you write the balanced equation for reaction, and are reasonably mathematical, you could use the ratio of atomic masses for N and H, suitably multiplied, to mentally check which reactant is limiting.

--- Cheers, -- Mr. d.