PTHH : `Zn + 2HCl \to ZnCl_2 + H_2`

`n_{H_2} = \frac{3,7185}{24,79} = 0,15` `mol`

`⇒ m_{H_2} = 0,15 . 2 = 0,3` `g`

`a.`

Theo phương trình : `n_{Zn} = n_{H_2} = 0,15` `mol`

`⇒ m_{Zn} = 0,15 . 65 = 9,75` `g`

`⇒ m_{Cu} = 17 - 9,75 = 7,25` `g`

`⇒ %Zn = \frac{9,75}{17} . 100% ≈ 57,35%`

    `%Cu = \frac{7,25}{17} . 100% ≈ 42,65%`

`b.`

Theo phương trình : `n_{HCl} = 2n_{H_2} = 0,3` `mol`

`⇒ m_{HCl} = 0,3 . 36,5 = 10,95` `g`

`⇒ m_{\text{dd HCl}} = 10,95 : 10% = 109,5` `g`

Theo phương trình : `n_{ZnCl_2} = n_{H_2} = 0,15` `mol`

`⇒ C%_{ZnCl_2} = \frac{0,15 . 136}{109,5 + 17 - 0,3} . 100% ≈ 16,165%`

$#Derix247$

`a,`

`n_{H_2} = V/{24,79} = {3,7185}/{24,79} = 0,15(mol)`

`Zn+2HCl->ZnCl_2+H_2`

`Cu+HCl->` $\text{Không phản ứng}$

`Theo\ pt:n_{Zn} = n_{H_2} = 0,15(mol)`

`-> m_{Zn} = n xx M = 0,15 xx 65 = 9,75(g)`

`-> m_{Cu} = 17 - 9,75 = 7,25(g)`

`-> %m_{Zn} = {9,75}/{17} xx 100 ~~ 57,35%`

`-> %m_{Cu} = 100% - 57,35% = 42,65%`

`b,`

`Theo\ pt:n_{ZnCl_2} = n_{H_2} = 0,15(mol)`

`m_{ZnCl_2} = n xx M = 0,15 xx 136 = 20,4\ (g)`

`Theo\ pt:n_{HCl} = 2n_{H_2} = 2 xx 0,15=0,3(mol)`

`m_{HCl} = n xx M = 0,3 xx 36,5 = 10,95(g)`

$\rm m_{dd\ HCl} = \dfrac{m_{ct}}{C\%} \times 100 = \dfrac{10,95}{10} \times 100 = 109,5\ (g)$

`m_{H_2} = n xx M = 0,15 xx 2 = 0,3\ (g)`

`m_{dd\ sau} = 17+109,5-0,3 = 126,2\ (g)`

`->` $\rm = C\%_{ ZnCl_2} = \dfrac{m_{ct}}{m_{dd}} \times 100 = \dfrac{20,4}{126,2} \times 100 \approx 16,165\%$