$a)$

$n_{H_2}=0.6(mol)$

Bảo toàn nguyên tố $H:$

$n_{H_2SO_4}=n_{H_2}=0.6(mol)$

Bảo toàn nguyên tố $S:$

$n_{SO_4}=n_{H_2SO_4}=0.6(mol)$

$m_{muối}=m_{\text{kim loại}}+m-{SO_4}=12.6+96 \times 0.6=70.2(gam)$

$n_R:n_{Al}=3:2=3a:2a$

Bảo toàn $e:$

$3a \times 2+ 3 \times 2a=2n_{H_2}=1.2(mol)$

`=>`$a=0.1$

`->`$M_R \times 3 \times 0.1+27 \times 2 \times 0.1=12.6$

`=>`$M_R=24$

Vậy $R$ là $magnesi(Mg)$

`a)`

BTNT H: `n_{H_2SO_4}=n_{H_2}={13,44}/{22,4}=0,6(mol)`

BTKL: `m_{m u ố i}=12,6+0,6.98-0,6.2=70,2(g)`

`b)`

Bte: `2n_R+3n_{Al}=2n_{H_2}=1,2(mol)`

`n_R:n_{Al}=3:2`

`->n_R=0,3(mol);n_{Al}=0,2(mol)`

`->M_R={12,6-0,2.27}/{0,3}=24(g//mol)`

`->R:Mg`