giúp em gấp vs anh chị ơi ;-;;;;

$a)$

`n_(H_2` `=(3,36)/(22,4)=0,15(mol)`

   `Mg+2HCl->MgCl_2+H_2`

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

`m_{Mg}=0,15.24=3,6(g)`

`m_{Cu}=16,4-3,6=12,8(g)`

$b)$

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

`n_{HCl(\text{bđ})}=0,3.125%=0,375(mol)`

`V_(ddHCl` `=(0,375)/(1)=0,375(l)=375(ml)`

`m_(ddHCl` `=375.1,08=405(g)`

Đáp án `+` Giải thích các bước giải:

$\rm a)$

$\rm n_{H_2} = \dfrac{3,36}{22,4} = 0,15(mol)$

$\rm PTHH: Mg + 2HCl \rightarrow MgCl_2 + H_2$
Theo PT: $\rm n_{Mg} = n_{H_2} = 0,15(mol)$
$\rm m_{Mg} = 0,15.24 = 3,6(g)$
$\rm m_{Cu} = 16,4 - 3,6 = 12,8(g)$
$\rm b)$
Theo PT: $\rm n_{HCl(pư)} = 2n_{H_2} = 0,3(mol)$
$\rm \Rightarrow n_{HCl(b đ)} = 0,3.(100+25)\% = 0,375(mol)$
$\rm \Rightarrow V_{ddHCl} = \dfrac{0,375}{1} = 0,375(l) = 375(ml)$
$\rm \Rightarrow m_{ddHCl} = 375.1,08 = 405(g)$