@$\text{Ly}$

$a)$

$n_{Zn}$ $=$ $\dfrac{26}{65}$ $=$ $0,4$ $mol$

PTHH: $Zn$ $+$ $2HCl$ `->` $ZnCl_2$ $+$ $H_2↑$

Theo PT, $n_{H_2}$ $=$ $n_{Zn}$ $=$ $0,4$ $mol$

$V_{H_2}$ $=$ $0,4.22,4=8,96$ $(l)$

$b)$

Theo PT, $n_{HCl}$ $=$ $2n_{Zn}$ $=$ $0,8$ $mol$

$n_{HCl}$ $=$ $0,8.36,5=29,2$ $(g)$

`C%_(HCl` $=$ $\dfrac{29,2}{50}$`.100%` $=$ $58,4$`%`

$c)$

Theo PT, $n_{ZnCl_2}$ $=$ $n_{Zn}$ $=$ $0,4$ $mol$

$m_{ZnCl_2}$ $=$ $0,4.136=54,4$ $(g)$

$m_{ddZnCl_2}$ $=$ $26+50-(0,4.2)=75,2$ $(g)$

`C%_(ZnCl_2` $=$ $\dfrac{54,4}{75,2}$`.100%` $≈$ $72,34$`%`

 

Đáp án `+` Giải thích các bước giải:

`a)` 

`n_(Zn) = m/M = (26)/(65) = 0,4` `(mol)`

`Zn+2HCl->ZnCl_2+H_2uparrow` 

Ta có : `n_(H_2) = n_(Zn)= 0,4` `(mol)`

`=>` `V_(H_2) = nxx22,4 = 0,4xx22,4 = 8,96` `(l)`

`b)`

Ta có : `n_(HCl) = 2n_(Zn)= 0,8` `(mol)`

`=>` `m_(HCl) = nxxM = 0,8xx26,5 = 29,2` `(g)`

`=>` `C%_(HCl) = (m_(ct))/(m_(dd)xx100%` `= (29,2)/(50)xx100%` `= 58,4%`

`c)`

Ta có : `n_(ZnCl_2) = n_(Zn)= 0,4` `(mol)`

`=>` `m_(ZnCl_2) = nxxM = 0,4xx136 = 54,4` `(g)`

`m_(dd sau pư) = 26+50-0,4xx2 = 75,2` `(g)`

`=>` `C%_(ZnCl_2) = (m_(ct))/(m_(dd)xx100%` `= (54,4)/(75,2)xx100%` `\approx` `72,34%`