Đáp án:

 5,6l

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
Mn{O_2} + 4HCl \to MnC{l_2} + C{l_2} + 2{H_2}O\\
nMn{O_2} = \dfrac{{21,75}}{{87}} = 0,25\,mol\\
nC{l_2} = nMn{O_2} = 0,25\,mol \Rightarrow VC{l_2} = 0,25 \times 22,4 = 5,6l\\
b)\\
C{l_2} + 2NaOH \to NaCl + NaClO + {H_2}O\\
m{\rm{dd}}NaOH = 1,28 \times 150 = 192g\\
nNaOH = \dfrac{{192 \times 15\% }}{{40}} = 0,72\,mol\\
m{\rm{ddA = 192 + 0,25}} \times {\rm{71 = 227,5 g}}\\
{\rm{nNaOH = 0,72 - 0,5 = 0,22}}\,{\rm{mol}}\\
{\rm{C\% NaOH = }}\dfrac{{0,22 \times 40}}{{227,5}} \times 100\%  = 3,87\% \\
C\% NaCl = \dfrac{{0,25 \times 58,5}}{{227,5}} \times 100\%  = 6,43\% \\
C\% NaClO = \dfrac{{0,25 \times 74,5}}{{227,5}} \times 100\%  = 8,19\% 
\end{array}\)