Stoichiometric calculation

Question

9.7 g H2 x 1 mol H_2 x _____=55g g NH3.

———-

2.0g H_2

N2 + 3H2 arrow 2NH3

Answer 1

Not really sure what your question is asking. It is written in a rather unusual fashion. But here is what I can offer.

Balanced equation:

N₂ + 3H₂ ==> 2NH₃

Given:

9.7 g H₂ x 1 mole H₂/2.0 g H₂ = 4.85 moles H₂

55 g NH₃ x 1 mole NH₃/17 g NH₃ = 3.24 moles NH₃

Calculated:

moles N₂ = 4.85 moles H₂ x 1 moles N₂/3 moles H₂ = 1.62 moles N₂

moles N₂ = 3.24 moles NH₃ x 1 mole N₂/2 moles NH₃ = 1.62 moles N₂

If you want grams of N2 it is 1.62 moles N2 x 28 g/mole = 45.4 g

Or by conservation of mass you have 9.7 g + x g = 55 g and x = g N2 = 45.3 g N2