Theoretical Yield Example Problem

This example problem demonstrates how to predict the amount of product formed by a given amount of reactants. This predicted quantity is the theoretical yield. Theoretical yield is the amount of product a reaction would produce if the reactants reacted completely.

Problem

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Theoretical Yield Worked Problem

By Todd Helmenstine

Given the reaction
Na₂S(aq) + 2 AgNO₃(aq) → Ag₂S(s) + 2 NaNO₃(aq)
How many grams of Ag₂S will form when 3.94 g of AgNO₃ and an excess of Na₂S are reacted together?

Solution

The key to solving this type of problem is to find the mole ratio between the product and the reactant.
Step 1 - Find the atomic weight of AgNO₃ and Ag₂S.
From the periodic table:
Atomic weight of Ag = 107.87 g
Atomic weight of N = 14 g
Atomic weight of O = 16 g
Atomic weight of S = 32.01 g
Atomic weight of AgNO₃ = (107.87 g) + (14.01 g) + 3(16.00 g)
Atomic weight of AgNO₃ = 107.87 g + 14.01 g + 48.00 g
Atomic weight of AgNO₃ = 169.88 g
Atomic weight of Ag₂S = 2(107.87 g) + 32.01 g
Atomic weight of Ag₂S = 215.74 g + 32.01 g
Atomic weight of Ag₂S = 247.75 g
Step 2 - Find mole ratio between product and reactant
The reaction formula gives the whole number of moles needed to complete and balance the reaction. For this reaction, two moles of AgNO₃ is needed to produce one mole of Ag₂S.
The mole ratio then is 1 mol Ag₂S/2 mol AgNO₃

_Step 3 Find amount of product produced.
The excess of Na₂S means all of the 3.94 g of AgNO₃ will be used to complete the reaction.
grams Ag₂S = 3.94 g AgNO₃ x 1 mol AgNO₃/169.88 g AgNO₃ x 1 mol Ag₂S/2 mol AgNO₃ x 247.75 g Ag₂S/1 mol Ag₂S
Note the units cancel out, leaving only grams Ag₂S
grams Ag₂S = 2.87 g Ag₂S

Answer

2.87 g of Ag₂S will be produced from 3.94 g of AgNO₃.