Đáp án:

$2Al + 3H_{2}SO_{4}→Al_{2}(SO_{4})_{3} + 3H_{2}↑$

a)

Ta có: $m_{H_{2}SO_{4}}=\dfrac{m_{dd} × \text{C%}}{\text{100%}}=\dfrac{98 × 30}{100}= 29,4 (g)$

⇒ $n_{H_{2}SO_{4}}=\dfrac{m}{M}=\dfrac{29,4}{98}= 0,3 (mol)$

Theo PT: $n_{H_{2}}=n_{H_{2}SO_{4}}= 0,3 (mol)$

⇒ $V_{H_{2}}=n × 22,4=0,3 × 22,4= 6,72 (l)$

b) Theo PT: $n_{Al_{2}(SO_{4})_{3}}=\dfrac{1}{3}n_{H_{2}SO_{4}}=\dfrac{1}{3} × 0,3= 0,1 (mol)$

⇒ $m_{Al_{2}(SO_{4})_{3}}=n × M=0,1 × 342= 34,2 (g)$

$Al_{2}(SO_{4})_{3}$: Nhôm sunfat

`#Ly`

$a)$

`m_(H_2SO_4` `=` `(m_{ddH_2SO_4}.C%)/(100)` `=` `(98.30)/(100)` `=` `29,4` $(g)$

`n_(H_2SO_4` `=` `m/M` `=` `(29,4)/(98)` `=` $0,3$ $(mol)$

PTHH: $2Al$ $+$ $3H_2SO_4$ `->` $Al_2(SO_4)_3$ $+$ $3H_2↑$

Theo PT, `n_(H_2` `=` `n_(H_2SO_4` `=` $0,3$ $(mol)$

`V_(H_2` `=` $n.22,4=0,3.22,4=6,72$ $(l)$

$b)$

Theo PT, `n_(Al_2(SO_4)_3` `=` `(1)/(3)n_(H_2SO_4` `=` $0,1$ $(mol)$

`m_(Al_2(SO_4)_3` `=` `n.M=0,1.342=34,2` $(g)$

`Al_2(SO_4)_3:` Nhôm sunfat