8.78 grams of nitrogen gas are allowed to react with 103 grams of hydrogen gas what’s is the Maximus mass of ammonium that can be formed

Answer 1

The first thing you should do in addressing this problem is to WRITE THE BALANCED EQUATION:

N₂(g) + 3H₂(g) ==> 2NH₃(g) ... balanced equation (note the ratio of 1 mol N₂ : 3 mol H₂)

Next, we will determine the limiting reactant by seeing which reactant produces the least amount of product:

For N₂: 8.78 g N₂ x 1 mol N₂/28 g x 2 mol NH₃/mol N₂ x 17 g NH₃/mol NH₃ = 10.7 g NH₃

For H₂: 103 g H₂ x 1 mol H₂/2 g x 2 mol NH₃/3 mol H₂ x 17 g NH₃/mol NH₃ = 584 g NH₃

LIMITING REACTANT IS N₂

Maximum amount of ammonium (NH₃) that can be formed = 10.7 g (to 3 significant figures)

Answer 2

8.78g N₂ / 1 mol/14.007g = 0.6268 mol N

103g H₂ / 1 mol/1.008g = 102.2 mol H

limiting reactant = N

Ammonium = NH₄ atomic mass = 14.007+4*1.008=18.039 g/mol

.6268 mol N * (1mol NH₄ /1mol N) *(18.039 g NH₄ /mol NH₄ ) = 11.3 g ammonium