Answer to Question #30682 in Chemistry for Kel

Question

If you have 84.0 grams of N2 and 12.0 grams of H2 for the reaction shown below, what is the liminting reactant?
N2 (g) + 3H2 (g)------> 2NH3 (g)

Answer 1

According to this reaction we will calculate what amount of H₂ is being needed for 84.0 grams of N₂.
n = m/M
where n - amount of moles
m - mass
M - molar mass
n(N₂) = 84/24 = 3.5 moles
n(H₂) = 3*n(N₂) = 3*3.5 = 10.5 moles
So the needed amount of H₂ is 10.5 moles. The real amount of H₂ in 12 grams is:
n(H₂) = 12/1 = 12 moles
The needed amount of H₂ is greater than the real one. So the amount of this element is limiting.