$25)$

$T+HCl$ cho khí.

`=>`$T$ chứa $Fe$ dư.

`=>`$Z$ chỉ có $FeCl_2$

Bảo toàn nguyên tố $Cl:$

$2n_{FeCl_2}=n_{HCl}$

`=>`$n_{FeCl_2}=\dfrac{0.36}{2}=0.18(mol)$

Bảo toàn nguyên tố $H:$

$n_{HCl}=2n_{H_2O}+2n_{H_2}$

`=>`$n_{H_2O}=\dfrac{0.36-2 \times 0.09}{2}=0.09(mol)$

Bảo toàn khối lượng:

$m+m_{HCl}=m_{FeCl_2}+m_T+m_{H_2}+m_{H_2O}$

`<=>`$m=0.18 \times (56+71)+16.56+0.09 \times 2+0.09 \times 18-36.5 \times 0.36$

`=>`$m=28.08(gam)$

`T : Cu , Fe` dư
BTNT `Cl`
`->` `1/2`$n_{HCl}$ `=` $n_{FeCl_2}$ `=0,36 . 1/2 = 0,18` mol
`BTNT` `H`
`->` $2n_{H_2O}$ `=` $n_{HCl}$ `-` $2n_{H_2}$
`->` $n_{H_2O}$ `=` $\dfrac{0,36 - 0,09 .2}{2}$ `=0,09` mol
`BTKL`
`-> m = 16,56 + 0,18 . 127 + 0,09 . 18 + 0,09 . 2 - 0,36 . 36,5 = 28,08g`