$n_{NO}=\dfrac{4.958}{24.79}=0.2(mol)$

$n_{NH_4^+}=a(mol)$

$n_{Al}=b(mol);n_{Mg}=c(mol)$

`->`$27b+24c=15$$(1)$

`-` Bảo toàn $e:$

$3b+2c=8a+0.2 \times 3$

`->`$-8a+3b+2c=0.6$$(2)$

$n_{NO_3^-}=n_{NH_4^+}+3n_{NO}=8a+0.6(mol)$

$m_{muối}=m_{Al+Mg}+m_{NO_3^-}+m_{NH_4NO_3}$

`->`$15+62(8a+0.6)+80a=109.8$

`=>`$a=0.1(mol)$$(3)$

`-` Từ $(1),(2),(3)$ suy ra:$b=0.2;c=0.4$

$n_{HNO_3,pu}=10n_{NH_4^+}+4n_{NO}=1.8(mol)$

`=>`$m_{dd,HNO_3}=\dfrac{ 63 \times 1.8 \times 115\% \times 100}{40}=326.025(gam)$

`-` Bảo toàn khối lượng:

$m_{X}=15+326.025-0.2 \times 30=335.025(gam)$

$\%C_{Al(NO_3)_3}=12.89\%$

$\%C_{Mg(NO_3)_2}=17.67\%$

$\%C_{NH_4NO_3}=2.39\%$

$\%C_{HNO_3}=5.08\%$

`n_{NO}={4,958}/{24,79}=0,2(mol)`

Đặt `n_{Al}=x(mol);n_{Mg}=y(mol)`

`->27x+24y=15(1)`

TH1: Không tạo `NH_4NO_3`

`Al^0->Al^{+3}+3e`

`Mg^0->Mg^{+2}+2e`

`N^{+5}+3e->N^{+2}`

Bảo toàn e: `3x+2y=3n_{NO}=0,6(2)`

`(1)(2)->` Âm, loại.

TH2: Tạo `NH_4NO_3`

`Al^0->Al^{+3}+3e`

`Mg^0->Mg^{+2}+2e`

`N^{+5}+8e->N^{-3}`

`N^{+5}+3e->N^{+2}`

Bảo toàn e: `n_{NH_4NO_3}={3x+2y-3n_{NO}}/8={3x+2y-0,6}/8(mol)`

Bảo toàn Mg,Al: `n_{Mg(NO_3)_2}=y(mol);n_{Al(NO_3)_3}=x(mol)`

`->213x+148y+10(3x+2y-0,6)=109,8(3)`

`(1)(3)->x=0,2;y=0,4`

`->n_{NH_4NO_3}=0,1(mol)`

Bảo toàn e: `n_{HNO_3}=10n_{NH_4NO_3}+4n_{NO}=1,8(mol)`

`->n_{HNO_3\ du}=1,8.15\%=0,27(mol)`

`m_{dd\ spu}=15+{(1,8+0,27).63}/{40\%}-0,2.30=335,025(g)`

\(\to\begin{cases} C\%_{Mg(NO_3)_2}=\dfrac{0,4.148}{335,025}.100\%≈17,67\%\\ C\%_{Al(NO_3)_3}=\dfrac{0,2.213}{335,025}.100\%≈12,72\%\\ C\%_{NH_4NO_3}=\dfrac{0,1.80}{335,025}.100\%≈2,39\%\\ C\%_{HNO_3\rm\ dư}=\dfrac{0,27.63}{335,025}.100\%≈5,08\%\end{cases}\)