$n_{NO}=a(mol);n_{N_2O}=b(mol)$

`->`$30a+44b=3.72$ $(1)$

$n_{N_2}=\dfrac{pV}{RT}=\dfrac{0.23 \times 3.2}{0.0821 \times (0+273)}=0.033(mol)$

$n_{khí}=\dfrac{pV}{RT}=\dfrac{1.1 \times 3.2}{0.0821 \times( 27.3+273)}=0.143(mol)$

`->`$a+b+0.033=0.143$ $(2)$

Từ $(1),(2)$ suy ra: $a=0.08;b=0.03$

$n_{Mg}=x(mol);n_{Zn}=y(mol);n_{Al}=z(mol)$

`->`$24x+65y+27z=7.5$ $(3)$

Bảo toàn $e:$

$2x+2y+3z=3a+8b=0.48$ $(4)$

Bảo toàn $e$ khi $A$ vào $KOH:$

$2y+3z=2n_{H_2}$

`=>`$n_{H_2}=\dfrac{2y+3z}{2}$

$m_{tăng}=m_{A}-m_{Mg}-m_{H_2}$

`->`$7.5-24x-2 \times \dfrac{2y+3z}{2}=5.82$ $(5)$

Từ $(3),(4),(5)$ suy ra: $x=\dfrac{3}{55};y=\dfrac{1569}{25850};z=\dfrac{43}{517}$

$\%m_{Mg}=17.45\%;\%m_{Zn}=52.60\%;\%m_{Al}=29.95\%$