3H2(g)+N2(g)→2NH3(g) 1.83 g H2 is allowed to react with 9.84 g N2, producing 3.00 g NH3. What is the theoretical yield in grams for this reaction under the given conditions?

Question

The answer in Value and unit.

Answer 1

3H2(g) + N2(g) ==> 2NH3(g) balanced equation

First, find limiting reactant:

moles H2 = 1.83 g x 1 mole/2 g = 0.915 moles H2

moles N2 = 9.84 g N2 x 1 mole/28 g = 0.351 moles N2

mole ratio of H2: N2 is 3:1, so H2 is limiting (0.915 is less than 3 x 0.351)

Theoretical yield of NH3 = 0.915 mol H2 x 2 mol NH3/3 mol H2 = 0.61 moles NH3

Theoretical mass of NH3 = 0.61 moles x 17 g/mole = 10.37 g

The percent yield, which you asked about in another question would be actual yield/theoretical yield (x100%) = 3.00 g/10.37 g (x100%) = 28.9%