Đáp án+Giải thích các bước giải:

`Mg + 2HCl \to MgCl_2 + H_2 `

`Fe + 2HCl \to FeCl_2 + H_2 `

`Zn + 2HCl \to ZnCl_2 + H_2 `

`a)`

`n_{H_2}=6,72÷22,4=0,3 mol`

Theo pt:

`n_{HCl}=2.nH2=2.0,3=0,6 mol`

`BTKL:`

`m_{Muối} =m_{KL}+m_{HCl}-m_{H_2}`

              `=14,5+0,6.36,5-0,3.2=35,8g`

`b)`

$CuO + H_2 \xrightarrow {t^0} Cu + H_2O $

Theo pt:

`n_{H_2O}=n_{H_2}=0,3 mol`

`BTKL:`

`m_{CuO.bđ}=m_{CR}+m_{H_2O}-m_{H_2}`

                `=23,2+18.0,3-2.0,3=28g`

a,

$n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol)$

$Mg+2HCl\to MgCl_2+H_2$

$Fe+2HCl\to FeCl_2+H_2$

$Zn+2HCl\to ZnCl_2+H_2$

Theo PTHH:

$n_{HCl}=2n_{H_2}=0,6(mol)$

BTKL:

$m_{\text{muối}}=14,5+0,6.36,5-0,3.2=35,8g$

b,

$CuO+H_2\xrightarrow{{t^o}} Cu+H_2O$

- Nếu chất rắn là $Cu$:

$n_{Cu}=\dfrac{23,2}{64}=0,3625(mol)$

$\to n_{H_2\text{pứ}}=n_{Cu}=0,3625(mol)>0,3$

$\to$ vô lí, loại 

- Nếu có $CuO$ dư 

$\to H_2$ hết

$\to n_{H_2O}=n_{H_2}=0,3(mol)$

BTKL:

$m_{CuO}=0,3.18+23,2-0,3.2=28g$