Đáp án:
\(\begin{array}{l}
{V_{C{l_2}}} = 2,8l\\
{m_{MnC{l_2}}} = 6,3g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
2KMn{O_4} + 16HCl \to 2KCl + 2MnC{l_2} + 5C{l_2} + 8{H_2}O\\
{n_{HCl}} = 0,4mol\\
\to {n_{C{l_2}}} = \dfrac{5}{{16}}{n_{HCl}} = 0,125mol\\
\to {V_{C{l_2}}} = 2,8l\\
\to {n_{MnC{l_2}}} = \dfrac{1}{8}{n_{HCl}} = 0,05mol\\
\to {m_{MnC{l_2}}} = 6,3g
\end{array}\)
