60.0 mL of 0.322 M potassium iodide are combined with 20.0 mL of 0.530 M lead () nitrate. What is the limiting reactant? How many grams of precipitate form?

Limiting reagent: Potassium iodide

Mass of the precipitate (PbI₂) is 4.453 g

Explanation:

We are given;

60.0 mL of 0.322 M potassium iodide 20.0 mL of 0.530 M lead () nitrate

We are required to identify the limiting reactant and determine the mass of the precipitate formed.

Step 1: Write the balanced equation for the reaction The balanced equation for the reaction between potassium iodide and lead (II) nitrate is given by;

2KI + Pb (NO₃) ₂ → 2KNO₃ + PbI₂ (s)

Step 2: Determine the number of moles of the reagents

Moles of KI

Moles = Molarity * volume

Moles of KI = 0.322 M * 0.060 L

= 0.01932 moles

Moles of KNO₃

Moles = 0.530 M * 0.020 L

= 0.0106 M

From the equation;

2 moles of KI reacts with 1 mole of Pb (NO) ₂ Therefore; 0.01932 moles of KI will require 0.00966 moles of Pb (NO₃) ₂ This means, KI is the limiting reagent while Pb (NO₃) ₂ is the excess reagent. Step 3: Determine the mass of the precipitate PbI₂

2 moles of KI reacts to produce 1 mole of PbI₂

Therefore;

Moles of PbI₂ = Moles of KI : 2

= 0.01932 moles : 2

= 0.00966 moles

But molar mass of PbI² is 461.01 g/mol

Therefore;

Mass of PbI₂ = 0.00966 moles * 461.01 g/mol

= 4.453 g

Therefore, the mass of the precipitate formed (Pbi₂) is 4.453 g