# Identify the type of reaction taking place in the following cases and write the balanced equation for the reaction(a) Potassium iodide reacts with lead nitrate to produce potassium nitrate and lead iodide(b) Zn reacts with silver nitrate to produce zinc nitrate and silver

Lead (II) Nitrate + Potassium Iodide → Lead (II) Iodide + Potassium Nitrate

In this reaction, both the reactants' exchange ions form two new products, hence, it is a double displacement reaction.

The symbol equation is as follows:

Pb(NO3)2 + KI → PbI2 + KNO3

The most obvious change we must make, when balancing this equation, is to increase the number of nitrate ions on the right-hand side of the equation. We can do this by placing a coefficient of 2 before the potassium nitrate:

Pb(NO3)2 + K I→ PbI2 + 2KNO3

In doing this we have upset the balance of potassium ions on each side of the equation. Again, we can fix this: we must simply place another coefficient of 2, this time before the potassium iodide:

Pb(NO3)2 + 2KI → PbI2 + 2KNO3

Similarly,

Zinc + Silver nitrate → Zinc nitrate + Silver

The balanced chemical equation  for the above reaction is given as :

Zn + 2AgNO3 Zn(NO3)2 + 2Ag.

In this reaction, zinc replaces silver in the final compound. Hence, it is a Displacement reaction.

It can be represented generically as: A + B-C → A-C + B .

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Updated on: 10-Oct-2022

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