Em tham khảo nha :

\(\begin{array}{l}
a)\\
2KCl{O_3} \xrightarrow{t^0} 2KCl + 3{O_2}\\
{n_{KCl}} = \dfrac{m}{M} = \dfrac{{44,7}}{{74,5}} = 0,6mol\\
hh\,A:KCl{O_3}(a\,mol),KCl(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,6\\
122,5a + 74,5b = 49,5
\end{array} \right.\\
 \Rightarrow a = 0,1;b = 0,5\\
{n_{{O_2}}} = \dfrac{3}{2}{n_{KCl{O_3}}} = 0,15mol\\
{V_{{O_2}}} = 0,15 \times 22,4 = 3,36l\\
b)\\
{m_{KCl{O_3}}} = 0,1 \times 122,5 = 12,25g\\
\% KCl{O_3} = \dfrac{{12,25}}{{49,5}} \times 100\%  = 24,75\% \\
\% KCl = 100 - 24,75 = 75,25\% \\
c)\\
4P + 5{O_2} \xrightarrow{t^0} 2{P_2}{O_5}\\
{n_P} = \dfrac{{3,72}}{{31}} = 0,12mol\\
\dfrac{{0,12}}{4} = \dfrac{{0,15}}{5} \Rightarrow \text{Phản ứng xảy ra hoàn toàn}\\
{n_{{P_2}{O_5}}} = \dfrac{{0,12}}{2} = 0,06mol\\
{m_{{P_2}{O_5}}} = 0,06 \times 142 = 8,52g\\
H = \dfrac{{7,1}}{{8,52}} \times 100\%  = 83,33\% 
\end{array}\)