Đáp án:

 \(Zn\) (kẽm)

\(V = 3,36{\text{ lít}}\)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(X + 2HCl\xrightarrow{{}}XC{l_2} + {H_2}\)

Ta có:

\({n_X} = \frac{{9,75}}{{{M_X}}};{n_{XC{l_2}}} = \frac{{20,4}}{{{M_X} + 2{M_{Cl}}}} = \frac{{20,4}}{{{M_X} + 35,5.2}} = \frac{{20,4}}{{{M_X} + 71}}\)

\( \to {n_X} = {n_{XC{l_{2}}}} \to \frac{{9,75}}{{{M_X}}} = \frac{{20,4}}{{{M_X} + 71}}\)

\( \to M_X=65 \to X:Zn\) 

\({n_{{H_2}}} = {n_{Zn}} = \frac{{9,75}}{{65}} = 0,15{\text{ mol}}\)

\( \to V = {V_{{H_2}}} = 0,15.22,4 = 3,36{\text{ lít}}\)

$PTHH:X+2HCl\to XCl_2+H_2\\ \Rightarrow n_X=n_{XCl_2}\\ \Rightarrow \dfrac{9,75}{M_X}=\dfrac{20,4}{M_X+71}\\ \Rightarrow M_X=65(g/mol)$

Vậy X là kẽm (Zn)

Theo PTHH ta có $n_{H_2}=n_{Zn}=\dfrac{9,75}{65}=0,15(mol)$

$\Rightarrow V=V_{H_2}=0,15.22,4=3,36(l)$