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)

N2 (g) + 3H2 (g)------> 2NH3 (g)

Expert's answer

According to this reaction we will calculate what amount of H_{2} is being needed for 84.0 grams of N_{2}.

n = m/M

where n - amount of moles

m - mass

M - molar mass

n(N_{2}) = 84/24 = 3.5 moles

n(H_{2}) = 3*n(N_{2}) = 3*3.5 = 10.5 moles

So the needed amount of H_{2} is 10.5 moles. The real amount of H_{2} in 12 grams is:

n(H_{2}) = 12/1 = 12 moles

The needed amount of H_{2} is greater than the real one. So the amount of this element is limiting.

n = m/M

where n - amount of moles

m - mass

M - molar mass

n(N

n(H

So the needed amount of H

n(H

The needed amount of H

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