Question

Lead ions can be precipitated from solution with nacl according to the following reaction. pb2 (aq) + 2nacl(aq) → pbcl2(s) + 2na (aq). When 135.8 g of nacl are added to a solution containing 195.7 g of pb2, a pbcl2 precipitate forms. The precipitate is filtered and dried and found to have a mass of 252.5 g. What is the percent yield of pbcl2?

Answer 1

To calculate the percent yield of PbCl2, you need to use the formula: Percent Yield = (Actual yield / Theoretical yield) x 100%. The actual yield is given as 252.5 g, and the theoretical yield can be calculated using the stoichiometry of the reaction.

Step-by-step explanation:

The percent yield of PbCl2 can be calculated using the formula:

Percent Yield = (Actual yield / Theoretical yield) x 100%

In this case, the actual yield is given as 252.5 g.

The theoretical yield can be calculated based on the stoichiometry of the reaction. From the balanced equation, we can see that 1 mol of PbCl2 is formed from 2 moles of NaCl. So, the moles of NaCl can be calculated using the given mass (135.8 g) and the molar mass of NaCl (58.44 g/mol).

Then, using the molar ratio, we can calculate the moles of PbCl2 formed. Finally, the theoretical yield can be calculated by multiplying the moles of PbCl2 by its molar mass (278.10 g/mol).

Once you have the actual yield and the theoretical yield, you can plug them into the percent yield formula to calculate the answer.