Đáp án:
1) Zn
2) 0,1l
Giải thích các bước giải:
\(\begin{array}{l}
1)\\
2M + nC{l_2} \to 2MC{l_n}\\
nC{l_2} = \frac{{6,72}}{{22,4}} = 0,3\,mol\\
= > nMC{l_n} = \frac{{0,6}}{n}mol\\
MMC{l_n} = \frac{{40,8}}{{\frac{{0,6}}{n}}} = 68n\\
= > MM = 32,5n\\
n = 2 = > MM = 65\\
= > M:Zn\\
2)\\
C{l_2} + 2NaOH \to NaCl + NaClO + {H_2}O\\
nC{l_2} = \frac{{1,12}}{{22,4}} = 0,05\,mol\\
= > nNaOH = 0,1\,mol\\
VNaOH = \frac{{0,1}}{1} = 0,1l
\end{array}\)