Determine the yield of N2+3H2=2NH3 if 100g of each reactant produces 100g of one of the products

Answer 1

This type of problem deals with limiting reagent, theoretical yield, and percent yield:

N₂ + 3H₂ -> 2NH₃ 100 grams of both Nitrogen and Hydrogen gas react to form 100 grams of NH₃

We can see from the balanced chemical equation that we need 1 molecule of nitrogen gas to react with 3 molecules of hydrogen gas to produce 2 molecules of NH₃ gas. Now this is the basis of a limiting reagent. One of the reactants (N₂ or H₂) will fully be used up before the other reactant. How much product is created before one of the reactants is used up is known as the theoretical yield, and the reactant that is used up completely is known as the limiting reagent.

For this problem to solve to find the limiting reagent, we need to see how much product 100 grams of each reactant can create (using stoichiometry).

100 grams N₂ --> 3.57 moles N₂ --> 7.14 moles NH₃

100 grams H₂ --> 49.5 moles H₂ --> 33.0 moles NH₃

When 7.14 moles of product is created, the nitrogen will be depleted. This means that nitrogen is the limiting reagent, and the theoretical yield is 7.14 moles of NH₃. The problem asked for percent yield given that 100 grams of product formed, so the theoretical yield is also 122 grams (instead of in moles).

Using the equation for percent yield = (Actual/Theoretical) *100

% yield =( 100 grams / 122 grams )*100 = 82.0%