Answer to Question #30682 in Other Chemistry for Kel
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)
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.