How many grams of hydrogen must react if the reaction needs to produce 63.5 grams of ammonia?

Answer 1

Assume all of the nitrogen reacts to form ammonia, and it is all in the gaseous state.

1 N2 + 3 H2 --> 2 NH3. This means there are two moles of ammonia for every mole of N2., or 2 x 22.4 = 44.8 Liters of ammonia at STP for every 28 g of N2

So 56.0 g N2 will yield 89.6 L of ammonia gas.

Your teacher probably wants the stoichiometry in the "railroad tracks", or "conversion factor" format where you start with the given, write a series of ratios that are all equal to 1, and multiply them together so all of the units drop out and you have the correct answer. This method will also help you catch your mistakes and solve many more difficult problems with ease (with a little practice).

In this case:

(56.0 g N2) x (1 mol N2/28g N2) X (2 mol NH3/1 mol N2) x

(22.4 L NH3 at STP / mol NH3) = 89.6 L of ammonia gas at STP.

Answer 2

For this question, you can just ignore the excess.

4H2 + N2 → 2NH4

4 mol H2 + 1 mol N2 → 2 mol NH4

6.96/4 = 1.74

6.96 mol H2 + 1.74 mol N2 → 3.48 mol NH4

3.48 mol NH4

3.48 mol NH4 x 18.05 g NH4/ 1 mol NH4 = 62.81 g NH4

Some nitrogen gas N2 will also be produced because there was an excess.

Answer 3

The reaction equation is N2 + 3 H2 -> 3 NH3 . The gram molecular mass of ammonia is 17.03, and the gram Atomic Mass of nitrogen is 14.0067. Therefore, the mass ratio of ammonia to the nitrogen used to make it is 17.03/14.067, and the mass of the ammonia produced from 50.0 g of nitrogen is 50.0 times the ratio or 60.5 g, to the justified number of significant digits.

Answer 4

0.94 grams of hydrogen

Answer 5

0.6 moles of N2 produce 1.2 moles of NH3

Answer 6

56.0 grams