Lời giải:\(\begin{array}{l}
{n_{KCl}} = 0,2 \to {n_{KCl{O_3}}} = 0,2{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {m_Y} = \frac{{14,9}}{{0,36315}} = 41,03{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \\
\to n_{{O_2}}^ \uparrow = \frac{{52,550 - 41,03}}{{32}} = 0,36
\end{array}\)
Vì cho X hoặc Y tác dụng với HCl thì khối lượng muối như nhau nên.Ta có ngay :
\(\begin{array}{l}
{m_X} = 52,55\left\{ \begin{array}{l}
KMn{O_4}:a\\
KCl{O_3}:0,2\\
Mn{O_2}:b
\end{array} \right. \to \left\{ \begin{array}{l}
KCl:a + 0,2\\
MnC{l_2}:a + b
\end{array} \right.\\
\to \left\{ \begin{array}{l}
74,5(a + 0,2) + 126(a + b) = 51,275\\
158a + 87b = 52,55 - 24,5
\end{array} \right.
\end{array}\)
\(\begin{array}{l}
\to \left\{ \begin{array}{l}
a = 0,15\\
b = 0,05
\end{array} \right.{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \\
{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 2KMn{O_4} \to {K_2}Mn{O_4} + Mn{O_2} + {O_2}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \\
\to H\% = \frac{{0,36 - 0,3}}{{0,075}} = 80\%
\end{array}\)
