Question

1.62 g H2 is allowed to react with 10.3 g N2, producing 1.09 g NH3

What is the theoretical yield in grams and the percentage yield for this reaction under the given conditions?

Answer 1

N2 + 3H2 → 2NH3

Moles H2 = mass / molar mass = 1.62 g / 2.016 g/mol
= 0.8036 mol
3 moles H2 react to produce 2 moles NH3
So moles NH2 possible = 2/3 x H2 = 0.5357 mol

Moles N2 = 10.3 g / 28.02 g/mol
= 0.3676 mol

1 mole N2 reacts to produce 2 moles NH

So moles NH3 possible = 2 x moles N2 = 0.7352 mol
H2 is the limiting reagent because the amount provided yields the least NH3

The theoretical yield of a reaction is the maximum possible product given the amounts of reagent provided. It occurs if all the limiting reagent is used up.

So in this case theoretical yield of NH3 = 0.5357 mol (amount produced from H2)

Mass NH3 (theoretical yiel) = molar mass x moles
= 17.034 g/mol x 0.5357 mol
= 9.13 g

% yield = actual yield / theoretical yield x 100/1
= 1.62 g / 9.13 g x 100/1
=17.7%