Đáp án:
14. D
15. B
16. A
Giải thích các bước giải:
\(\begin{array}{l}
14.\\
Mn{O_2} + 4HCl \to MnC{l_2} + 2{H_2}O + C{l_2}\\
{n_{Mn{O_2}}} = 0,1mol\\
\to {n_{C{l_2}}} = {n_{Mn{O_2}}} = 0,1mol\\
\to {V_{C{l_2}}} = 2,24l
\end{array}\)
\(\begin{array}{l}
15.\\
Mn{O_2} + 4HCl \to MnC{l_2} + 2{H_2}O + C{l_2}\\
{n_{C{l_2}}} = 0,2mol\\
\to {n_{Mn{O_2}}} = {n_{C{l_2}}} = 0,2mol\\
\to {m_{Mn{O_2}}} = 17,4g
\end{array}\)
\(\begin{array}{l}
16.\\
2KMn{O_4} + 16HCl \to 2KCl + 5C{l_2} + 8{H_2}O + 2MnC{l_2}\\
{n_{C{l_2}}} = 0,1mol\\
\to {n_{KMn{O_4}}} = \dfrac{2}{5}{n_{C{l_2}}} = 0,04mol\\
\to {m_{KMn{O_4}}} = 6,32g
\end{array}\)
