Đáp án:

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

$CuO + H_{2}\xrightarrow{t^o}Cu + H_{2}O$ (2)

a) Ta có: $m_{H_{2}SO_{4}}=\dfrac{m_{dd} × \text{C%}}{\text{100%}}=\dfrac{67 × 20}{100}= 13,4 (g)$

⇒ $n_{H_{2}SO_{4}}=\dfrac{m}{M}=\dfrac{13,4}{98}≈ 0,1 (mol)$

Theo PT (1): $n_{H_{2}}=n_{H_{2}SO_{4}}= 0,1 (mol)$

⇒ $V_{H_{2}}=n × 22,4=0,1 × 22,4= 2,24 (l)$

b) Theo PT (1): $n_{Al_{2}(SO_{4})_{3}}=\dfrac{1}{3}n_{H_{2}SO_{4}}=\dfrac{1}{3} × 0,1= \dfrac{1}{30} (mol)$

⇒ $m_{Al_{2}(SO_{4})_{3}}=n × M=\dfrac{1}{30} × 342= 11,4 (g)$

c) Theo PT (2): $n_{H_{2}O}=n_{H_{2}}= 0,1 (mol)$

⇒ $m_{H_{2}O}=n × M=0,1 × 18= 1,8 (g)$

`#Ly`

$a)$

$m_{H_2SO_4}$ $=$ `(m_{ddH_2SO_4}.C%)/(100)` $=$ $\dfrac{67.20}{100}$ $=$ $13,4$ $(g)$

$n_{H_2SO_4}$ $=$ $\dfrac{m}{M}$ $=$ $\dfrac{13,4}{98}$ $≈$ $0,14$ $(mol)$

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

Theo PT, $n_{H_2}$ $=$ $n_{H_2SO_4}$ $=$ $0,14$ $(mol)$

$V_{H_2}$ $=$ $n.22,4=0,14.22,4=3,136$ $(l)$

$b)$

Theo PT, $n_{Al_2(SO_4)_3}$ $=$ $\dfrac{1}{3}n_{H_2}$ $=$ $0,04$ $(mol)$

$m_{Al_2(SO_4)_3}$ $=$ $n.M=0,04.342=13,68$ $(g)$

$c)$ 

PTHH: $CuO$ $+$ $H_2$ $\xrightarrow[]{t^o}$ $Cu$ $+$ $H_2O$

Theo PT trên, $n_{H_2O}$ $=$ $n_{H_2}$ $=$ $0,14$ $(mol)$

$m_{H_2O}$ $=$ $n.M=0,14.18=2,52$ $(g)$