$8.$

$n_{H_2}=0.6(mol)$

Bảo toàn nguyên tố $H:$

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

Bảo toàn nguyên tố $S:$

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

$m_{muối}=m_{KL}+96n_{H_2SO_4}$

`<=>`$93.6=m+96 \times 0.6$

`=>`$m=36(g)$

$9.$

$n_{H_2SO_4}=1.3 \times 0.5=0.65(mol)$

Bảo toàn nguyên tố $H:$

$n_{H_2}=n_{H_2SO_4}=0.65(mol)$

`=>`$V=16.1135(L)$

$n_{Fe}=a(mol);n_{Al}=b(mol)$

`->`$\begin{cases} 56a+27b=19.3\\2a+3b=0.65 \times 2 \text{ (Theo bảo toàn e) } \end{cases}$

$\begin{cases} a=0.2\\b=0.3 \end{cases}$

Bảo toàn nguyên tố $Al:$

$n_{Al_2(SO_4)_3}=n_{Al}=0.3(mol)$

`=>`$m=102.6(gam)$

`8)`

BTNT H: `n_{H_2SO_4}=n_{H_2}={14,874}/{24,79}=0,6(mol)`

BTKL: `m_X+m_{H_2SO_4}=m_{m u o i}+m_{H_2}`

`->m=m_X=93,6+0,6.2-0,6.98=36(g)`

`9)`

BTNT H: `n_{H_2}=n_{H_2SO_4}=1,3.0,5=0,65(mol)`

`->V=0,65.24,79=16,1135(L)`

Đặt `n_{Fe}=x(mol);n_{Al}=y(mol)`

`56x+27y=19,3(1)`

BTe: `2x+3y=2n_{H_2}=1,3(2)`

`(1)(2)->x=0,2;y=0,3`

BTNT Al: `n_{Al_2(SO_4)_3}=1/2y=0,15(mol)`

`->m=0,15.342=51,3(g)`