`#Ly`

$a)$

PTHH:

$2Al$ $+$ $6HCl$ `->` $2AlCl_3$ $+$ $3H_2↑$

$Mg$ $+$ $2HCl$ `->` $MgCl_2$ $+$ $H_2↑$

$AlCl_3$ $+$ $3NaOH$ `->` $Al(OH)_3↓$ $+$ $3NaCl$

$MgCl_2$ $+$ $2NaOH$ `->` $Mg(OH)_2$ $+$ $2NaCl$

$2Al(OH)_3$ $\xrightarrow[]{t^o}$ $Al_2O_3$ $+$ $3H_2O$

$Mg(OH)_2$ $\xrightarrow[]{t^o}$ $MgO$ $+$ $H_2O$

`n_(H_2` `=` `(2,24)/(22,4)=0,1` $(mol)$

Coi `n_(Al` `=` `x` $(mol)$, `n_(Mg` `=` `y` $(mol)$

Ta có:

`n_(H_2` `=` `(3)/(2)n_(Al` `=` `1,5x` $(mol)$

Và `n_(H_2` `=` `n_(Mg` `=` `y` $(mol)$

`∑``n_(H_2` `=` `1,5x+y=0,1` $(mol)$

Bảo toàn $Al:$ `n_(Al_2O_3` `=` `(1)/(2)n_(Al` `=` `0,5x` $(mol)$

Bảo toàn $Mg:$ `n_(MgO` `=` `n_(Mg` `=` `y` $(mol)$

`m_(CR` `=` `51x+40y=3,7` $(g)$

Ta có hệ PT:

`1,5x+y=0,1`

`51x+40y=3,7`

`=>` $\begin{cases} x=0,033\\y=0,05 \end{cases}$

`m_(Al` `=` `0,033.27=0,891` $(g)$

`m_(Mg` `=` `0,05.24=1,2` $(g)$

`m_(hh` `=` `0,891+1,2=2,091` $(g)$

$b)$

Ta có:

`n_(HCl` `=` `3n_(Al` `=` `0,099` $(mol)$

`n_(HCl` `=` `2n_(Mg` `=` `0,1` $(mol)$

`n_(KOH` `=` `0,1.0,1=0,01` $(mol)$

$KOH$ $+$ $HCl$ `->` $KCl$ $+$ $H_2O$

`->` `n_(HCl` `=` `n_(KOH` `=` `0,01` $(mol)$

`∑``n_(HCl` `=` `0,099+0,1+0,01=0,209` $(mol)$

`C_MHCl` `=` `(0,209)/(0,15)≈1,39` $(M)$

$c)$

Ta có: `n_(KCl` `=` `n_(KOH` `=` `0,01` $(mol)$

`C_MKCl` `=` `(0,01)/(0,1)=0,1` $(M)$