Đáp án:
\(\begin{array}{l}
a){V_{C{l_2}}} = 11,2l\\
b)\\
{m_{KMn{O_4}(dư)}} = 31,6g\\
{m_{MnC{l_2}}} = 25,2g\\
{m_{KCl}} = 14,9g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
2KMn{O_4} + 16HCl \to 2KCl + 5C{l_2} + 2MnC{l_2} + 8{H_2}O\\
a)\\
{n_{HCl}} = 1,6mol\\
{n_{KMn{O_4}}} = 0,4mol\\
\to \dfrac{{{n_{KMn{O_4}}}}}{2} > \dfrac{{{n_{HCl}}}}{{16}} \to {n_{KMn{O_4}}}dư\\
\to {n_{C{l_2}}} = \frac{5}{{16}}{n_{HCl}} = 0,5mol\\
\to {V_{C{l_2}}} = 11,2l\\
b)\\
{n_{KMn{O_4}}} = {n_{MnC{l_2}}} = {n_{KCl}} = \dfrac{1}{8}{n_{HCl}} = 0,2mol\\
\to {n_{KMn{O_4}(dư)}} = 0,2mol\\
\to {m_{KMn{O_4}(dư)}} = 31,6g\\
\to {m_{MnC{l_2}}} = 25,2g\\
\to {m_{KCl}} = 14,9g
\end{array}\)
