Đáp án:
\(\begin{array}{l}
a)\\
{V_{C{l_2}}} = 0,56l\\
b)\\
{m_{NaClO}} = 1,8625g\\
{m_{NaCl}} = 1,4625g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Mn{O_2} + 4HCl \to MnC{l_2} + C{l_2} + 2{H_2}O\\
{n_{Mn{O_2}}} = \dfrac{m}{M} = \dfrac{{2,175}}{{87}} = 0,025mol\\
{n_{C{l_2}}} = {n_{Mn{O_2}}} = 0,025mol\\
{V_{C{l_2}}} = n \times 22,4 = 0,025 \times 22,4 = 0,56l\\
b)\\
C{l_2} + 2NaOH \to NaClO + NaCl + {H_2}O\\
{n_{NaClO}} = {n_{NaCl}} = {n_{C{l_2}}} = 0,025mol\\
{m_{NaClO}} = n \times M = 0,025 \times 74,5 = 1,8625g\\
{m_{NaCl}} = n \times M = 0,025 \times 58,5 = 1,4625g
\end{array}\)