$n_{hh}$ `=`$\dfrac{1,7353}{24,79}$ `=0,07` mol
Đặt $n_{NO}$ `=x` mol , $n_{N_2O}$ `=y` mol
`-> x + y = 0,07` `(I)`
$m_{hh}$ `=30x + 44y = 0,07 . 18 .2 = 2,52(II)`
`(I),(II)` `->` `x=0,04` và `y=0,03`
Đặt $n_{Al}$ `=a` mol và $n_{Mg}$ `=b` mol
`-> 27a + 24b = 3,6g``(III)`
Bte
`-> ` $3n_{Al}$ `+` $2n_{Mg}$ `=` $3n_{NO}$ `+` $8n_{N_2O}$
`-> 3a + 2b = 3. 0,04 + 8 . 0,03 = 0,36 ( IV)`
`(III)` và `(IV)`
`-> a = 0,08 ` vầ `b= 0,06`
`=>` $m_{Al}$ `=0,08 . 27 = 2,16g`

`=>` $m_{Mg}$ `=0,06 . 24 = 1,44g`

`M_{khí}=18.2=36(g//mol)`

`->{n_{NO}}/{n_{N_2O}}={44-36}/{36-30}=4/3`

`n_{NO}+n_{N_2O}={1,7353}/{24,79}=0,07(mol)`

`->n_{NO}=0,04(mol);n_{N_2O}=0,03(mol)`

`Hh:Al:x(mol);Mg:y(mol)`

`27x+24y=3,6(1)`

Bte: `3x+2y=3n_{NO}+8n_{N_2O}=0,36(2)`

`(1)(2)->x=0,08;y=0,06`

`m_{Al}=0,08.27=2,16(g)`

`m_{Mg}=0,06.24=1,44(g)`