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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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