Em tham khảo nha :
\(\begin{array}{l}
a)\\
2Pb{(N{O_3})_2} \to 4N{O_2} + {O_2} + 2PbO\\
{n_{Pb{{(N{O_3})}_2}}} = \dfrac{{66,22}}{{331}} = 0,2mol\\
\text{Theo đề ta có :}\\
{m_{hh}} = 66,22 - 55,4 = 10,82g\\
{m_{N{O_2}}} + {m_{{O_2}}} = 10,82g\\
\Rightarrow 46{n_{N{O_2}}} + 32{n_{{O_2}}} = 10,82\\
\Rightarrow 46 \times 4{n_{{O_2}}} + 32{n_{{O_2}}} = 10,82\\
\Rightarrow {n_{{O_2}}} = \dfrac{{10,82}}{{216}} = 0,05mol\\
{n_{Pb{{(N{O_3})}_2}_{pu}}} = 2{n_{{O_2}}} = 0,1mol\\
H = \dfrac{{0,1}}{{0,2}} \times 100\% = 50\% \\
b)\\
{n_{N{O_2}}} = 4{n_{{O_2}}} = 0,2mol\\
{V_{{O_2}}} = 0,05 \times 22,4 = 1,12l\\
{V_{N{O_2}}} = 0,2 \times 22,4 = 4,48l
\end{array}\)
