Đáp án:
a) 2,8l
b) 0,71M
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2KMn{O_4} + 16HCl \to 2KCl + 2MnC{l_2} + 5C{l_2} + 8{H_2}O\\
5 \times |2C{l^ - } \to C{l_2} + 2e\\
2 \times |M{n^{ + 7}} + 5e \to M{n^{ + 2}}\\
{n_{KMn{O_4}}} = \dfrac{{7,9}}{{158}} = 0,05\,mol\\
{n_{C{l_2}}} = 0,05 \times \frac{5}{2} = 0,125\,mol\\
{V_{C{l_2}}} = 0,125 \times 22,4 = 2,8l \Rightarrow V = 2,8\\
b)\\
2NaOH + C{l_2} \to NaCl + NaClO + {H_2}O\\
{n_{NaOH}} = 2{n_{C{l_2}}} = 0,25\,mol\\
{C_M}NaOH = \dfrac{{0,25}}{{0,35}} \approx 0,71M \Rightarrow a = 0,71
\end{array}\)
