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

`⇒27a+56b=5,54`(1*)

`n_{H_{2}}=\frac{3,584}{22,4}=0,16(mol)`

`2Al+3H_{2}SO_{4}→Al_{2}(SO_{4})_{3}+3H_{2}(1)`

`a`                                                                   `1,5a`           / `mol`

`Fe+H_{2}SO_{4}→FeSO_{4}+H_{2}(2)`

`b`                                             `b`         / `mol`

`⇒n_{H_{2}}=1,5a+b=0,16`(2*)

Từ (1*) và (2*) ⇒$\left \{ {{a=0,06} \atop {b=0,07}} \right.$ 

`⇒%m_{Al}=\frac{0,06.27}{5,54}.100=29,24%`

`⇒%m_{Fe}=100%-29,24%=70,76%`

`Theo phương trình `(1);(2)⇒n_{H_{2}SO_{4}}=n_{H_{2}}=0,16(mol)`

`⇒V_{ddH_{2}SO_{4}}=\frac{n_{H_{2}SO_{4}}}{C_{M_{ddH_{2}SO_{4}}}}=\frac{0,16}{1,5}=\frac{8}{75}` ( lít )

`a)`

$\rm 2Al+3H_2SO_4 \rightarrow{} Al_2(SO_4)_3+3H_2$

$\rm Fe+H_2SO_4\rightarrow{}FeSO_4+H_2$

$\rm ∑n_{H_2}=\dfrac{3.584}{22.4}=0.16(mol)$

Gọi $\rm n_{Al}=x (mol); n_{Fe}=y(mol)$

Ta có hệ sau:

`{(27x+56y=5.54),(3/2x+y=0.16):}`

$\rm ⇒ x=0.06, y=0.07$

$\rm \%m_{Al}=\dfrac{0.06 \times 27}{5.54}\times 100\% \approx 29.24\%$

$\rm \% m_{Fe}=\dfrac{0.07 \times 56}{5.54}\times 100\% \approx 70.76\%$

`b)`

PT: $\rm ∑n_{H_2SO_4}=∑n_{H_2}=0.16(mol)$

$\rm \rightarrow V_{dd \kern3pt H_2SO_4}=\dfrac{0.16}{1.5} \approx 0.11(L)=110(mL)$

#TS