a,

$n_{HCl}=0,2.2=0,4(mol)$

$n_{H_2}=\dfrac{3,36}{22,4}=0,15(mol)$

$2Al+6HCl\to 2AlCl_3+3H_2$

$CuO+2HCl\to CuCl_2+H_2O$

$\Rightarrow n_{Al}=\dfrac{2}{3}n_{H_2}=0,1(mol)$

Ta có: $3n_{Al}+2n_{CuO}=n_{HCl}$

$\Rightarrow n_{CuO}=\dfrac{0,4-0,1.3}{2}=0,05(mol)$

$m_{Al}=0,1.27=2,7g$

$m_{CuO}=0,05.80=4g$

$\to m=2,7+4=6,7g$

b,

$V_{dd}=0,2l$

$n_{AlCl_3}=n_{Al}=0,1(mol)$

$\to C_{M_{AlCl_3}}=\dfrac{0,1}{0,2}=0,5M$

$n_{CuCl_2}=n_{CuO}=0,05(mol)$

$\to C_{M_{CuCl_2}}=\dfrac{0,05}{0,2}=0,25M$

Giải thích các bước giải:

Vì `V_{HCl}=200(ml)=0,2(l)`

`→n_{HCl}=0,2.2=0,4(mol)`

`-n_{H_2}=\frac{3,36}{22,4}=0,15(mol)`

Phương trình hóa học :

`CuO+2HCl→CuCl_2+H_2O(1)`

`0,05`   `←0,1`   `→0,05`                `(mol)`    

`2Al+6HCl→2AlCl_3+3H_2↑(2)`

`0,1`         `0,3`        `0,1`   `←`    `0,15`        `(mol)`

`→m_{Al}=0,1.27=2,7(g)`

`-n_{HCl}(1)=∑n_{HCl}-n_{HCl}(2)=0,4-0,3=0,1(mol)`

`→m_{CuO}=0,05.80=4(g)`

`⇒m=m_{CuO}+m_{Al}=4+2,7=6,7(g)`

`b.`

Dung dịch `A` gồm `:CuCl_2;AlCl_3`

`-V_{dd.sau.pứ}=V_{dd.ban.đầu}=V_{HCl}=0,2(l)` 

`→CM_{CuCl_2}=\frac{0,05}{0,2}=0,25(M)`

`→CM_{AlCl_3}=\frac{0,1}{0,2}=0,5(M)`

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