Question

Lead nitrate on heating decomposes to give:
$2Pb(N{O}_3{)}_2\to 2PbO+4N{O}_2+O_3$
What is the total volume of gases produced when 13.24 g of lead nitrate is heated? All gases are measured at STP.
$(Pb=207;N=14;O=16)$

• A. 2.24 litres
• B. 1.12 litres
• C. 2 litres
• D. 1.24 litres

Answer 1

Answer: (A)

2.24 litres

Lead nitrate on heating decomposition to gives:

$2Pb(N{O}_3{)}_2⟶2PbO+4N{O}_2+O_3$

$\left(g\right)\left(g\right)\left(g\right)\left(g\right)\left(g\right)$

Molar mass of $Pb(N{O}_3{)}_2=207+2(4)+6(16)=3Hg$

so, $N{O}_2$ and $O_2$ gases are released.

Given weight of $Pb(N{O}_3{)}_2=13.24g$

Number of moles of $Pb(N{O}_3{)}_2={\left(\frac{33.1}{13.24}\right)}^{-1}=(25{)}^{-1}=\frac{1}{25}$

$\frac{1}{25}$ moles of $Pb(N{O}_3{)}_2$ gives $\frac{1}{2}\left(\frac{4}{25}\right)$ moles of $N{O}_2$ and $\frac{1}{25}\left(\frac{1}{25}\right)$ moles of $N{O}_2$ and $O_2$

$\Rightarrow \frac{1}{2}\left(\frac{5}{25}\right)$ moles of gases are liberated.

We know, 1 mole gas $22.4L$ Volume.

$\frac{1}{2}\left(\frac{5}{25}\right)$ moles $\Rightarrow 22.4\times \left(\frac{5}{25}\right)\frac{1}{2}$

Total Volume of ases $=2.24$ Litres.