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
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.