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Answer to Question #30682 in Other Chemistry for Kel

Question #30682
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)
Expert's answer
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.

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