Answer :
Answer:
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
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
- For potassium iodide (KI)
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
- For lead nitrate [Pb(NO₃)₂]
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 [tex]\frac{19.32}{2}[/tex] moles of Pb(NO₃)₂
[tex]\frac{19.32}{2} = 9.66[/tex]
∴ 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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