Đáp án:

\(\begin{array}{l}
{m_{KMn{O_4}}} = 6,32g\\
{V_{HCl}} = 0,56l
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
2KMn{O_4} \to {K_2}Mn{O_4} + Mn{O_2} + {O_2}\\
2KMn{O_4} + 16HCl \to 5C{l_2} + 2KCl + 2MnC{l_2} + 8{H_2}O\\
{K_2}Mn{O_4} + 8HCl \to 2C{l_2} + 4{H_2}O + 2KCl + MnC{l_2}\\
Mn{O_2} + 4HCl \to MnC{l_2} + C{l_2} + 2{H_2}O\\
{m_{{O_2}}} = {m_g} = 0,32g\\
{n_{{O_2}}} = \dfrac{{0,32}}{{32}} = 0,01mol\\
{n_{C{l_2}}} = \dfrac{{1,792}}{{22,4}} = 0,08mol\\
BT\,electron:\\
5{n_{KMn{O_4}}} = 2{n_{C{l_2}}} + 4{n_{{O_2}}}\\
 \Rightarrow {n_{KMn{O_4}}} = \dfrac{{0,08 \times 2 + 0,04}}{5} = 0,04mol\\
{m_{KMn{O_4}}} = 0,04 \times 158 = 6,32g\\
BT\,Clo:\\
{n_{HCl}} = {n_{KCl}} + 2{n_{MnC{l_2}}} + 2{n_{C{l_2}}}\\
 \Rightarrow {n_{HCl}} = 0,28mol\\
{V_{HCl}} = \dfrac{{0,28}}{{0,5}} = 0,56l
\end{array}\)