Đáp án:
C
Giải thích các bước giải:
\(\begin{array}{l}
KClO + 2HCl \to KCl + C{l_2} + {H_2}O\\
KCl{O_2} + 4HCl \to KCl + 2C{l_2} + 2{H_2}O\\
KCl{O_3} + 6HCl \to KCl + 3C{l_2} + 3{H_2}O\\
nC{l_2} = \frac{{8,064}}{{22,4}} = 0,36mol\\
TheoPT:nHCl = 2nC{l_2} = 0,72\,mol\\
n{H_2}O = nC{l_2} = 0,36\,mol\\
BTKL\\
m + mHCl = mKCl + mC{l_2} + m{H_2}O\\
= > m = mKCl + mC{l_2} + m{H_2}O - mHCl = 19,17g
\end{array}\)