Question

For the reaction of 60.0 mL of 0.322 mol/L potassium iodide and 20.0 mL of 0.530 mol/L lead(Il) nitrate, determine the % yield of this reaction if 4.29g of precipitate is collected.

EXTENSION for Q#8: Determine the concentration of excess reagent remaining!

Answer 1

The percent yield of the reaction is 48.2%.

Step-by-step explanation:

To determine the percent yield of the reaction, we need to calculate the theoretical yield of lead(II) iodide first. From the balanced chemical equation, we know that the stoichiometric ratio between lead(II) nitrate and lead(II) iodide is 1:1. This means that for every mole of lead(II) nitrate, we will produce one mole of lead(II) iodide.

First, we calculate the moles of potassium iodide and lead(II) nitrate used:

Moles of KI = 0.060 L * 0.322 mol/L = 0.01932 mol

Moles of Pb(NO3)2 = 0.020 L * 0.530 mol/L = 0.0106 mol

Since the stoichiometric ratio is 1:1, we can see that the limiting reactant is potassium iodide because it is used in a larger quantity than lead(II) nitrate.

Theoretical yield of PbI2 = Moles of limiting reactant * molar mass of PbI2

= 0.01932 mol * 461.01 g/mol = 8.898 g

Now, we can calculate the percent yield:

Percent yield = (Actual yield / Theoretical yield) * 100%

= (4.29 g / 8.898 g) * 100% = 48.2%