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?
1
Expert's answer
2016-09-12T06:12:03-0400
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%
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