\(\begin{array}{l} {n_{KCl}} = 0,2 \to {n_{KCl{O_3}}} = 0,2{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {m_Y} = \frac{{14,9}}{{0,36315}} = 41,03{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \\ \to n_{{O_2}}^ \uparrow = \frac{{52,550 - 41,03}}{{32}} = 0,36 \end{array}\)

Vì cho X hoặc Y tác dụng với HCl thì khối lượng muối như nhau nên.Ta có ngay :

\(\begin{array}{l} {m_X} = 52,55\left\{ \begin{array}{l} KMn{O_4}:a\\ KCl{O_3}:0,2\\ Mn{O_2}:b \end{array} \right. \to \left\{ \begin{array}{l} KCl:a + 0,2\\ MnC{l_2}:a + b \end{array} \right.\\ \to \left\{ \begin{array}{l} 74,5(a + 0,2) + 126(a + b) = 51,275\\ 158a + 87b = 52,55 - 24,5 \end{array} \right. \end{array}\)

\(\begin{array}{l} \to \left\{ \begin{array}{l} a = 0,15\\ b = 0,05 \end{array} \right.{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \\ {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 2KMn{O_4} \to {K_2}Mn{O_4} + Mn{O_2} + {O_2}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \\ \to H\% = \frac{{0,36 - 0,3}}{{0,075}} = 80\% \end{array}\)