A metal Nitrate 'A' on heating gives yellowish brown coloured metal oxide along with brown gas 'B' and a colourless gas 'C'. Aqueous solution of 'A' on reaction with potassium iodide forms a yollow precipitate of compound 'D'.Identify A,B,C,D.Also identify the type of both the reactions.

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Solution

This reaction is a decomposition reaction,The metal nitrate (A) is Lead nitrate (Pb(NO₃)₂) On heating it produces Lead oxide, Nitrogen dioxide (B), and oxygen gas (C)
The yellowish-brown colored metal oxide is lead nitrate (Pb(NO₃)₂)
The brown gas is nitrogen dioxide(B)
The colorless gas is oxygen gas (C)
$2 P b {(N O_{3})}_{2} (s) ⟶ 2 P b O (s) + 4 N O_{2} (g) + O_{2} (g)$
Here, lead nitrate is decomposed to give lead oxide, nitrogen dioxide, and oxygen gas.
This reaction is a decomposition reaction, where one reactant breaks down into two or more products.
Aqueous lead nitrate on reaction with potassium iodide mixture of lead iodide and potassium nitrate forms.
$P b {(N O_{3})}_{2} (a q) + 2 K I (s) ⟶ P b I_{2} (s) + 2 K N O_{3} (s)$
Here, lead nitrate is reacting with potassium iodide to form lead iodide and potassium nitrate, an exchange of ions takes place.
This reaction is a double displacement reaction where the exchange of ions between the compound takes place.
Alloy-made using lead is used for soldering purposes. Therefore,
A = (Pb(NO₃)₂)
B = NO₂
C = O₂
D = PbI + KNO₃ (mixture of compounds)

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