$a,$

$n_{HCl}=\dfrac{18,25}{36,5}=0,5(mol)$

$n_{H_2SO_4}=\dfrac{13,72}{98}=0,14(mol)$

$n_{H_2}=\dfrac{8,736}{22,4}=0,39(mol)$

$Mg+2HCl→MgCl_2+H_2$

$Mg+H_2SO_4→MgSO_4+H_2$

$2Al+3H_2SO_4→Al_2(SO_4)_3+3H_2$

$2Al+6HCl→2AlCl_3+3H_2$

Theo pt: $n_{H_2}=\dfrac{n_{HCl}}{2}+n_{H_2SO_4}=0,39(mol)$ ⇒ Axit hết

Bảo toàn $Cl,S:$

$n_{Cl}(muối)=n_{HCl}=0,5(mol)$

$n_{SO_4}(muối)=n_{H_2SO_4}=0,14(mol)$

$m_{muối}=m_{KL}+m_{Cl}+m_{SO_4}=7,74+0,5×35,5+0,14×96=38,93(g)$

$b,$

$MgCl_2+2NaOH→Mg(OH)_2↓+2NaCl$

$MgSO_4+2NaOH→Mg(OH)_2↓+Na_2SO_4$

$AlCl_3+3NaOH→Al(OH)_3↓+3NaCl$

$Al_2(SO_4)_3+6NaOH→2Al(OH)_3↓+3Na_2SO_4$

Theo pt $n_{NaOH}=n_{Cl}+2n_{SO_4}=0,5+2×0,14=0,78(mol)$

$a=0,78×40=31,2(g)$

Đáp án + Giải thích các bước giải:

 `Mg + H_2SO_4 → MgSO_4 + H_2` 

`Mg + 2HCl → MgCl_2 + H_2` 

`2Al + 6HCl → 2AlCl_3 + 3H_2` 

`2Al+ 3H_2SO_4 → Al_2(SO_4)_3 + 3H_2` 

`a)` 

`n_{H_2} = {8,736}/{22,4} = 0,39(mol)` 

$BTNT.H$ 

`n_{H_2} = n_{H_2SO_4} + 1/2n_{HCl}` 

`= {13,72}/{98} + {18,25}/{36,5} :2 = 0,39(mol)`

`→` Phản ứng vừa đủ

$BTKL$:

`m_{text{muối}} = m_{kl} + m_{text{axit}} - m_{H_2}`

`= 7,74 + 18,25 + 13,72 - 0,39 .2 = 38,93(g)` 

`b)` 

`3NaOH + AlCl_3 → Al(OH)_3 + 3NaCl` 

`Al_2(SO_4)_3 + 6NaOH → 2Al(OH)_3 + 3Na_2SO_4` 

`MgCl_2 + 2NaOH → Mg(OH)_2 + 2NaCl`

`MgSO_4 + 2NaOH → Mg(OH)_2 + Na_2SO_4` 

$BTNT.Cl$ 

`n_{NaCl} = n_{HCl} = {18,25}/{36,5} `

`=0,5(mol)` 

$BTNT.S$ 

`n_{Na_2SO_4} = n_{H_2SO_4} = {13,72}/{98}`

`= 0,14(mol)` 

$BTNT.Na$ 

`n_{NaOH} = n_{NaCl} + 2n_{Na_2SO_4}`

`= 0,5 + 0,14 . 2 = 0,78(mol)` 

`a = 0,78 . 40 = 31,2(g)`

`text{#lazuli}`