\(\begin{array}{l}
a)\\
Dat\,KL\,la\,R\,co\,hoa\,tri\,la\,n\\
2R+nH_2SO_4\to R_2(SO_4)_n+nH_2\\
n_{H_2}=\frac{1,12}{22,4}=0,05(mol)\\
Theo\,PT:\,n_R.n=2n_{H_2}=0,1\\
\to \frac{1,2n}{M_R}=0,1\\
\to M_R=12n\\
\to n=2;M_R=24\\
\to R:\,magie\,(Mg)\\
b)\\
Mg+H_2SO_4\to MgSO_4+H_2\\
Theo\,PT:\,n_{H_2SO_4}=n_{H_2}=0,05(mol)\\
\to m_{dd\,H_2SO_4}=\frac{0,05.98}{9,8\%}=50(g)\\
c)\\
Theo\,PT:\,n_{MgSO_4}=n_{H_2}=0,05(mol)\\
\to C\%_{MgSO_4}=\frac{0,05.120}{1,2+50-0,05.2}.100\%\approx 11,74\%
\end{array}\)

Gọi kim loại là $X$

$a)$

$nH_2$ = $\frac{1,12}{22,4}$ = $0,05$ $mol$

$2X$ + $H_2SO_4$ → $X_2(SO_4)_n$ + $nH_2$

$\frac{0,1}{n}$                                        ← $0,05$

⇒ $nX$ = $\frac{0,1}{n}$  = $\frac{1,2}{X}$ 

⇒ $X$ = $12n$ 

⇒ $n$ = $2$ ⇒ $X$ là $Mg$ 

$b)$

$Mg$ + $H_2SO_4$ → $MgSO_4$ + $H_2$ 

                $0,05$                                 $0,05$

⇒ $nH_2SO_4$ = $0,05$ mol

⇒ $mH_2SO_4$ = $0,05$ ×$9,8$ = $4,9$g

⇒ $mddH_2SO_4$ = $4,9$ : $9,8$ % = $50$g

$c)$

$Mg$ + $H_2SO_4$ → $MgSO_4$ + $H_2$ 

                $0,05$             $0,05$          $0,05$

⇒ $nMgSO_4$ = $0,05$ mol

⇒ $mMgSO_4$ = $0,05$ × $120$ = $6$g

mddspu = $50$ + $1,2$ = $51,2$ g

C% $MgSO_4$ = $\frac{6}{51,2}$ .100% = $11,71$ %