Đáp án:

`a. Zn + 2HCl -> ZnCl_2 + H_2`

`b.`

`m_{Zn}= 16,25 (g)`

`-> %m_{Zn}= 81,25 (%)`

`m_{Cu}= 3,75 (g)`

`-> %m_{Cu}= 18,75 (%)`

`c. C%_{ZnCl_2} = 15,76%`

Giải thích các bước giải:

`a.`

`Zn + 2HCl -> ZnCl_2 + H_2`

`Cu` không tác dụng được với `HCl`

`n_{H_2} = (5,6)/(22,4) = 0,25 (mol)`

Theo phương trình: `n_{Zn} = n_{ZnCl_2} = n_{H_2} = 0,25 (mol)`

`b.`

`m_{Zn} = 0,25 . 65 = 16,25 (g)`

`-> %m_{Zn} = (16,25 . 100)/20 = 81,25 (%)`

`m_{Cu} = 20 - 16,25 = 3,75 (g)`

`-> %m_{Cu} = 100 - 81,25 = 18,75 (%)`

`c.`

`m_{ZnCl_2} = 0,25 . (65 + 35,5 . 2) = 34 (g)`

`m_{ddsaupu} = m_{Zn} + m_{ddHCl} - m_{H_2}`

`-> m_{ddsaupu} = 16,25 + 200 - 0,25 . 2 = 215,75 (g)`

`-> C%_{ZnCl_2} = (34 . 100)/(215,75) = 15,76 (%)`

$n_{H_2}=\dfrac{5,6}{22,4}=0,25(mol)\\ a,PTHH:Zn+2HCl\to ZnCl_2+H_2\\ b,n_{Zn}=n_{H_2}=0,25(mol)\\ \Rightarrow m_{Zn}=65.0,25=16,25(g)\\ \Rightarrow m_{Cu}=20-16,25=3,75(g)\\ \Rightarrow \%_{Zn}=\dfrac{16,25}{20}.100\%=81,25\%\\ \Rightarrow \%_{Cu}=100\%-81,25\%=18,75\%\\ c,n_{ZnCl_2}=n_{H_2}=0,25(mol)\\ \Rightarrow m_{CT_{ZnCl_2}}=0,25.136=34(g)\\ \Rightarrow C\%_{ZnCl_2}=\dfrac{34}{16,25+200-0,25.2}.100\%\approx 15,76\%$