Đáp án:
1,6M
0,8M
Giải thích các bước giải:
\(\begin{array}{l}
Mn{O_2} + 4HCl \to MnC{l_2} + C{l_2} + 2{H_2}O\\
nMn{O_2} = \frac{{69,6}}{{87}} = 0,8\,mol\\
= > nC{l_2} = 0,8mol\\
nNaOH = 0,5 \times 4 = 2mol\\
2NaOH + C{l_2} \to NaCl + NaClO + {H_2}O\\
\frac{2}{2} > \frac{{0,8}}{1}
\end{array}\)
=> NaOH dư
\(\begin{array}{l}
nNaOH = 2 - 0,8 \times 2 = 0,4\,mol\\
nNaCl = nNaClO = 0,8\,mol\\
CMNaCl = CMNaClO = \frac{{0,8}}{{0,5}} = 1,6M\\
CMNaOH = \frac{{0,4}}{{0,5}} = 0,8M
\end{array}\)