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?

The limiting reactant is potassium iodide (KI) and the mass of precipitate (PbI₂) formed is 4453.36 g

From the question,

We are to determine the limiting reactant in the reaction between the given potassium iodide and lead nitrate

First, we will write a balanced chemical equation for the reaction

The balanced chemical equation for the reaction is

2KI + Pb(NO₃)₂ → 2KNO₃ + PbI₂

This means

2 moles of KI will react with 1 mole of Pb(NO₃)₂ to produce 2 moles of KNO₃ and 1 mole of PbI₂

To determine the limiting reactant,

First, we will determine the number of moles of each reactant present

Volume = 60.0 mL

Concentration = 0.322 M

From the formula

Number of moles = Concentration × Volume

∴ Number of moles of KI present = 0.322 × 60.0

Number of moles of KI present = 19.32 moles

Volume = 20.0 mL

Concentration = 0.530 M

∴ Number of moles of Pb(NO₃)₂ present = 0.530 × 20.0

Number of moles of Pb(NO₃)₂ present = 10.60 moles

Now, from the balanced chemical equation

2 moles of KI is required to react with 1 mole of Pb(NO₃)₂

Then,

19.32 moles of KI will react with moles of Pb(NO₃)₂

∴ 19.32 moles of KI will react with 9.66 moles of Pb(NO₃)₂

The number of moles of Pb(NO₃)₂ present is 10.60 moles.

Since this is more than the quantity that reacted (9.66 moles), then Pb(NO₃)₂ is the excess reactant and KI is the limiting reactant.

∴ KI is the limiting reactant

To determine how many grams of the precipitate formed,

The precipitate formed is lead iodide (PbI₂)

That is, we will determine the mass of lead iodide formed

From the balanced equation for the reaction,

2 moles of KI will react with 1 mole of Pb(NO₃)₂ to produce 2 moles of KNO₃ and 1 mole of PbI₂

Then,

19.32 moles of KI will react with 9.66 moles of Pb(NO₃)₂ to produce 19.32 moles of KNO₃ and 9.66 moles of PbI₂

Therefore, 9.66 moles of PbI₂ was formed during the reaction.

Now, for the mass of PbI₂ formed

From the formula

Mass = Number of moles × Molar mass

Number of moles of PbI₂ = 9.66 moles

Molar mass of PbI₂ = 461.01 g/mol

∴ Mass of PbI₂ formed = 9.66 × 461.01

Mass of PbI₂ formed = 4453.3566 g

Mass of PbI₂ formed ≅ 4453.36 g

Hence, the limiting reactant is potassium iodide (KI) and the mass of precipitate (PbI₂) formed is 4453.36 g

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Limiting reagent: Potassium iodide

Mass of the precipitate (PbI₂) is 4.453 g

We are given;

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

2KI + Pb(NO₃)₂ → 2KNO₃ + PbI₂(s)

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 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

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