a,
Xét TN2:
Gọi x, y là số mol $Al$, $Al_2O_3$
$\Rightarrow 27x+102y=6,45$ (1)
$n_{AlCl_3}=x+2y$ (bảo toàn Al)
$n_{AlCl_3}=0,15(mol)$
$\Rightarrow x+2y=0,15$ (2)
(1)(2)$\Rightarrow x=y=0,05$
Xét TN1:
$n_{Al(NO_3)_3}=x+2y=0,15(mol)$
$\Rightarrow m_{Al(NO_3)_3}=0,15.213=31,95g<32,7g$
Vậy B chứa $Al(NO_3)_3$, $NH_4NO_3$
b,
$n_{NH_4NO_3}=\dfrac{32,7-31,95}{80}=0,009375(mol)$
Bảo toàn e: $3n_{Al}=8n_{NH_4NO_3}+3n_{NO}$
$\Rightarrow n_{NO}=0,025(mol)$
$\to V=0,56l$
