Question 3.4.3: Consider the reaction of Pb(NO3)2 with NaCl to form PbCl2 an......
Consider the reaction of Pb(NO_{3})_{2} with NaCl to form PbCl_{2} and NaNO_{3}. If 24.2 g Pb(NO_{3})_{2} is reacted with excess NaCl and 17.3 g of PbCl_{2} is ultimately isolated, what is the percent yield for the reaction?
You are asked to calculate the percent yield for a reaction.
You are given the mass of the limiting reactant and the experimental yield for one of the products in the reaction.
Step 1. Write a balanced chemical equation for the reaction.
Pb(NO_{3})_{2} + 2 NaCl → PbCl_{2} + 2 NaNO_{3}
Step 2. Use the amount of limiting reactant to calculate the theoretical yield of PbCl_{2}.
24.2 g Pb\left(\text{NO}_{3}\right) _{2}\left(\frac{1\text{ mol Pb}\left(\text{NO}_{3}\right)_{2} }{331.2\text{ g}} \right) \left(\frac{1\text{ mol PbCl}_{2}}{1\text{ mol Pb}\left(\text{NO}_{3}\right)_{2}} \right) \left(\frac{278.1\text{ g}}{1\text{ mol PbCl}_{2}} \right) = 20.3 g PbCl_{2}
Step 3. Use the experimental yield (17.3 g PbCl_{2}) and the theoretical yield (20.3 g PbCl_{2}) to calculate the percent yield for the reaction.
\frac{17.3\text{ g PbCl}_{2}}{20.3\text{ g PbCl}_{2}} \times 100\% =85.2\%Is your answer reasonable? The theoretical yield is greater than the experimental yield, so the percent yield is less than 100%.