Đáp án + Giải thích các bước giải:

a) PTHH:
$Zn + 2HCl \longrightarrow ZnCl_2 + H_2\uparrow$
$Mg + 2HCl \longrightarrow MgCl_2 + H_2 \uparrow$

 b) $m_{Zn} = \dfrac{40,4 . 64,36\%}{100\%} = 26(g)$

$\Rightarrow n_{Zn} =\dfrac{m}{M} = \dfrac{26}{65} = 0,4(\text{mol})$
$m_{Mg} = 40,4 - 26 = 14,4(g)$

$\Rightarrow n_{Mg} = \dfrac{m}{M} = \dfrac{14,4}{24} = 0,6(\text{mol})$

Theo PTHH:

$n_{H_2} = n_{Zn} + n_{Mg} = 0,4 + 0,6 = 1(\text{mol})$
$\Rightarrow V_{H_2} = n . 22,4 = 1 . 22,4 = 22,4(l)$

c) Theo PTHH:
$n_{ZnCl_2} = n_{Zn} = 0,4(\text{mol})$
$n_{MgCl_2} = n_{Mg} = 0,6(\text{mol})$

$\Rightarrow m_{ZnCl_2} =n . M = 0,4 . (65 + 2 . 35,5) = 0,4 . 136 = 54,4(g)$

$m_{MgCl_2} = n . M = 0,6 . (24 + 2 . 35,5) = 0,6 . 95 = 57(g)$
$m_{\text{muối}} = 54,4 + 57 = 111,4(g)$

Đáp án:

 `↓`

Giải thích các bước giải:

Ta có :

`m_{Zn}=(64,36)/100xx40,4~~26(g)`

`=>m_{Mg}=40,4-26=14,4(g)`

`a)`

`Zn+2HCl->ZnCl_2+H_2`

`Mg+2HCl->MgCl_2+H_2`

`b)`

`\sum n_{H_2}=n_{Zn}+n_{Mg}=26/65+(14,4)/24=0,4+0,6=1(mol)`

`=> \sum V_{H_2}=1xx22,4=22,4(l)`

`c)`

`n_{ZnCl_2}=n_{Zn}=0,4(mol)`

`n_{MgCl_2}=n_{Mg}=0,6(mol)`

`m_{ZnCl_2}=0,4xx136=54,4(g)`

`m_{MgCl_2}=0,6xx95=57(g)`

`=>m_{text{Muối}}=57+54,4=111,4(g)`

`text{# Yor}`