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$n_{Al}:n_{Mg}=3a:2a$

`->`$27 \cdot 3a+24 \cdot 2a=12,9$

`->`$a=0,1$

`-` Bảo toàn $e:$

$3n_{Al}+2n_{Mg}=10n_{N_2}+8n_{N_2O}$

`->`$10n_{N_2}+8n_{N_2O}=1,3$$(1)$

$n_{khí}=0,15(mol)$

`->`$n_{N_2O}+ n_{N_2}=0,15$$(2)$

`-` Từ $(1),(2)$ suy ra: $n_{N_2}=0,05(mol);n_{N_2O}=0,1(mol)$

$m_{khí}=0,05 \cdot 28+0,1 \cdot 44=5,8(g)$

`->`$M_{khí}=\dfrac{5,8}{0,15}=\dfrac{116}{3}(g/mol)$

`=>`$a=\dfrac{116}{3}:2=\dfrac{58}{3}$

Có `27n_{Al}+24n_{Mg}=12,9`

Mà `n_{Al}:n_{Mg}=3:2`

`->n_{Al}=0,3(mol);n_{Mg}=0,2(mol)`

Đặt `n_{N_2}=x(mol);n_{N_2O}=y(mol)`

`->x+y={3,36}/{22,4}=0,15(1)`

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

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

`2N^{+5}+10e->N_2^0`

`N^{+5}+4e->N^{+1}`

Bảo toàn electron: `3n_{Al}+2n_{Mg}=10n_{N_2}+4n_{N_2O}`

`->10x+4y=3.0,3+2.0,2=1,3(2)`

`(1)(2)->x=7/{60};y=1/{30}`

`->M_{hh\ khí}={7/{60}.28+1/{30}.44}/{0,15}={284}/9(g//mol)`

`->a={{284}/9}/2\approx 15,78`