Sửa `46,1->4,61`

`a)`

`Hh:Mg:x(mol);Fe:y(mol);Zn:z(mol)`

`->24x+56y+65z=4,61(1)`

`Mg+2HCl->MgCl_2+H_2`

`Fe+2HCl->FeCl_2+H_2`

`Zn+2HCl->ZnCl_2+H_2`

`->n_{H_2(Mg)}=x(mol);n_{H_2(Fe)}=y(mol)`

`->2x=y(2)`

`\sum n_{H_2}=x+y+z={17,92}/{22,4}=0,08(3)`

`(1)(2)(3)->x=0,01;y=0,02;z=0,05`

`\%m_{Mg}={0,01.24}/{4,61}.100\%=5,21\%`

`\%m_{Fe}={0,02.56}/{4,61}.100\%=24,3\%`

`\%m_{Zn}=100-24,3-5,21=70,49\%`

`b)`

Theo pt:

`n_{HCl}=2n_{H_2}=0,16(mol)`

`m_{dd\ HCl}={0,16.36,5}/{25\%}=23,36(g)`

`V_{dd\ HCl}={23,36}/{1,2}=19,467(mL)`

`c)`

`m_{dd\ sau}=4,61+23,36-0,08.2=27,81(g)`

Theo pt:

`n_{MgCl_2}=0,01(mol);n_{FeCl_2}=0,02(mol);n_{ZnCl_2}=0,05(mol)`

`C\%_{MgCl_2}={0,01.95}/{27,81}.100\%=3,42\%`

`C\%_{FeCl_2}={0,02.127}/{27,81}.100\%=9,13\%`

`C\%_{ZnCl_2}={0,05.136}/{27,81}.100\%=24,45\%`