Ta có: \({M_{{X_2}}} = 3,1696.22,4 = 71 \to {M_X} = 35,5 \to X:{\text{ Cl}}\)
Vậy khí đó là \(C{l_2}\)
\({H_2} + C{l_2}\xrightarrow{{}}2HCl\)
\(Cu + C{l_2}\xrightarrow{{}}CuC{l_2}\)
\(2Fe + 3C{l_2}\xrightarrow{{}}2FeC{l_3}\)
\(2NaOH + C{l_2}\xrightarrow{{}}NaCl + NaClO + {H_2}O\)
\({H_2}O + C{l_2}\overset {} \leftrightarrows HCl + HClO\)