Em tham khảo nha :
\(\begin{array}{l}
2Pb{(N{O_3})_2} \to 4N{O_2} + {O_2} + 2PbO\\
{n_B} = \dfrac{{5,6}}{{22,4}} = 0,25mol\\
hh:{O_2}(a\,mol),N{O_2}(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,25\\
4a - b = 0
\end{array} \right.\\
\Rightarrow a = 0,05;b = 0,2\\
{n_{Pb{{(N{O_3})}_2}}} = 2{n_{{O_2}}} = 0,1mol\\
{m_{Pb{{(N{O_3})}_2}}} = 0,1 \times 332 = 33,2g\\
H = \dfrac{{33,2}}{{200}} \times 100\% = 16,6\%
\end{array}\)