$n_{Al}=\dfrac{12.42}{27}=0.46(mol)$

$n_{Y}=\dfrac{1.344}{22.4}=0.06(mol)$

$n_{N_2O}=a(mol);n_{N_2}=b(mol)$

`->`$a+b=0.06$ $(1)$

$M_Y=18 \times 2=36(g/mol)$

`->`$m_Y=0.06 \times 36=2.16(gam)$

`->`$44a+28b=2.16$$(2)$

Từ $(1),(2)$ suy ra : $a=b=0.03$

Bảo toàn $e:$

$3n_{Al}=8n_{N_2O}+10n_{N_2}+8n_{NH_4^+}$

`=>`$n_{NH_4^+}=\dfrac{0.46 \times 3-8 \times 0.03 -10 \times 0.03}{8}=0.105(mol)$

Ta có các bán phản ứng:

$10H^++\mathop{2N}\limits^{+5}O_3^- +8e^-\to \mathop{N_2}\limits^{+1}O$

$12H^++\mathop{2N}\limits^{+5}O_3^- +10e^-\to \mathop{N_2}\limits^{0}$

$10H^++\mathop{N}\limits^{+5}O_3^- +8e^-\to \mathop{N}\limits^{-3}H_4^+$

`=>`$n_{H^+}=10n_{N_2O}+12n_{N_2}+10n_{NH_4^+}$

$=10 \times 0.03+12 \times 0.03+10 \times 0.105=1.71(mol)$

Mà $n_{HNO_3}=n_{H^+}=1.71(mol)$

nAl=0.46(mol);

gọi nN2O=a,nN2=b

a+b=0.06

44a+28b=0.06*2*18=2.16

=>a=b=0.03

bte: 3nAl=8nN2O+10nN2+10nNH4+ =>nNH4+=0.105

btH =>nHNO3=nH+=10nN2O+12nN2+10nNH4+=1.71(mol)