1.21g H2 is allowed to react with 9.51g N2, producing 2.32g NH3. what is the theoretical yield for the reaction?

Answer 1

Write a correctly balanced equation:

3H₂(g) + N₂(g) ==> 2NH₃(g)

moles H₂ present = 1.21 g x 1 mole H₂/2.00 g = 0.605 moles H₂

moles N₂ present = 9.51 g N₂ x 1 mole N₂/28.00 g = 0.3396 moles N₂

Find which reactant is limiting:

For H₂ you get 0.605 moles H₂ x 2 moles NH₃/3 moles H₂ = 0.403 moles NH₃

For N₂ you get 0.3396 moles N₂ x 2 moles NH₃/1 mole N₂ = 0.679 moles NH₃

H₂ is limiting

Theoretical yield of NH3 = 0.403 moles NH3 (based on H2 being limiting) x 17 g/mole = 6.85 g NH3

% yield (not asked for) = 2.32 g/6.85 g (x100%) = 33.9% yield