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Question

Lead(II) nitrate on heating with an aqueous solution of potassium iodide, gives a yellow precipitate. Write a balanced equation for the reaction and name the type of reaction.


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Solution

1. Type of reaction

Combination reaction: Reaction in which two or more reactants combine to form a single product. A+BC

Decomposition reaction: Reaction in which reactant is broken down into two or more products in the presence of heat, light or electricity. AB+C

Displacement reaction: Reaction in which one of the higher reactive element displaces less reactive element from its solution. A+BCAC+B ( Here A is more reactive than B)

Double displacement reaction: Reaction in which ions of both the reactants interchange their place and form new compounds. AB+CDAD+CB

2. Reaction:

Pb(NO3)2+KIheatPbI2+KNO3

Lead(II) nitrate Potassium iodide Lead iodide Potassium nitrate

(Yellow)

On heating lead(II) nitrate with potassium iodide double displacement reaction takes place both Pb and K interchange their position.

3. Balancing steps:

Balancing steps:
Step 1: Write the unbalanced equation.
Pb(NO3)2+KIheatPbI2+KNO3

Step 2: Write down the number of atoms of each elements present on both reactant as well as product side.

Element Reactant side Product side
Pb 1 1
N 2 1
O 6 3

K 1 1

I 1 1

Step 3: Add coefficients to balance the number of atoms on both sides.
i. Balance N and O at product side by adding 2 before KNO3.
Pb(NO3)2+KIheatPbI2+2KNO3

ii. Balance K and I at reactant side by adding 2 before KI.

Pb(NO3)2+2KIheatPbI2+2KNO3

vi. Balanced equation is:
Pb(NO3)2(aq)+2KI(aq)heatPbI2(s)+2KNO3(aq)
Lead(II) nitrate Potassium iodide Lead iodide Potassium nitrate

(Yellow)

So it is a double displacement reaction.


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