Gọi x, y là số mol $Al$, $Al_2O_3$.
$\Rightarrow 27x+102y=4,2$ $(1)$
$n_{N_2O}=\dfrac{0,672}{22,4}=0,03(mol)$
Bảo toàn e: $3n_{Al}=8n_{N_2O}$
$\Rightarrow 3x=0,03.8$ $(2)$
$(1)(2)\Rightarrow x=0,08; y=0,02$
Bảo toàn Al:
$n_{Al(NO_3)_3}=n_{Al}+2n_{Al_2O_3}=0,12(mol)$
$\Rightarrow m_{Al(NO_3)_3}=0,12.213=25,56g$
Bảo toàn N:
$n_{HNO_3}=3n_{Al(NO_3)_3}+2n_{N_2O}=0,42(mol)$
$\Rightarrow m_{dd HNO_3}=0,42.63:10\%=264,6g$