1)`m_{text[rắn không tan]} = m_{Cu} = 0,64g`

Gọi `x, y` lần lượt là số mol của `Mg, Fe`

`=> 24x + 56y + 0,64 = 4,64(1)`

`Mg + 2HCl -> MgCl_2 + H_2`

`Fe + 2HCl -> FeCl_2 + H_2`

Theo PT

`n_{H_2} = n_{Mg} + n_{Fe} = x + y = (0,2)/(2) = 0,1` mol `(2)`

`(1),(2) => x = 0,05; y = 0,05`

`m_{Mg} = 0,05 . 24 = 1,2g`

`m_{Fe} = 0,05 . 56 = 2,8g`

2)

Theo PTHH

`n_{HCl} = 2n_{H_2} = 2 . 0,1 = 0,2` mol

`=> m_{dd HCl} = (0,2 . 36,5)/(10%) = 73g`

$1)\\ Mg+2HCl\xrightarrow\ MgCl_2+H_2\\ Fe+2HCl\xrightarrow\ FeCl_2+H_2\\ n_{Mg}:a(mol)\ ; n_{Fe}:b(mol)\\ \xrightarrow\ 24a+56b=4,64-0,64\ (1)\\ n_{H_2}=\dfrac{0,2}{2}=0,1(mol)\\ \sum n_{H_2}=n_{Mg}+n_{Fe}\\ \xrightarrow\ a+b=0,1\ (2)\\ (1)(2)\xrightarrow\ a=b=0,05(mol)\\ m_{Cu}=0,64(g)\\ m_{Mg}=0,05.24=1,2(g)\\ m_{Fe}=0,05.56=2,8(g)\\ 2)\\ n_{HCl}=2n_{H_2}=0,2(mol)\\ m_{dd\ HCl}=\dfrac{0,2.36,5}{10\%}=73(g)$