Lead nitrate of heating decomposes to give
$2Pb(N{O}_3{)}_2\to 2PbO+4N{O}_2+O_2$
What is the total volume of gases produced when 13.24 g of lead nitrate is heated? All gases are measured at STP.

$(Pb=207;N=14;O=16)$

The correct option is A 2.24 litres
$2Pb(N{O}_3{)}_2\to 2PbO+4N{O}_2+O_2$

in this reaction 2 mol of lead nitrate gives 4 mol of $N{O}_2$ and 1 mol of $O_2$

and in this question we are use $\frac{13.24}{331}=$0.04 mol of lead nitrate

so the question total mol produced $N{O}_2$ is 0.08 and $O_2$ is 0.02

total mol of gas is produced is 0.08+0.02 =0.1 mol because $PbO$ is solid

so the total volume of gases is $0.1\times 22.4$ liters =2.24 liters

so the answer is D