Đáp án:
a) 2,688l
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Mn{O_2} + 4HCl \to MnC{l_2} + C{l_2} + 2{H_2}O\\
nMn{O_2} = \dfrac{{10,44}}{{87}} = 0,12\,mol\\
nC{l_2} = nMn{O_2} = 0,12\,mol \Rightarrow VC{l_2} = 0,12 \times 22,4 = 2,688l\\
b)\\
C{l_2} + 2NaOH \to NaCl + NaClO + {H_2}O\\
nNaOH = \dfrac{{12}}{{40}} = 0,3\,mol\\
m{\rm{dd}}NaOH = 1,28 \times 150 = 192g\\
m{\rm{ddA = 192 + 0,12}} \times {\rm{71 = 200,52 g}}\\
{\rm{nNaOH = 0,3 - 0,24 = 0,06}}\,{\rm{mol}}\\
{\rm{C\% NaOH = }}\dfrac{{0,06 \times 40}}{{200,52}} \times 100\% = 1,2\% \\
C\% NaCl = \dfrac{{0,12 \times 58,5}}{{200,52}} \times 100\% = 3,5\% \\
C\% NaClO = \dfrac{{0,12 \times 74,5}}{{200,52}} \times 100\% = 4,46\%
\end{array}\)