a,
$n_{Al}=\dfrac{13,5}{27}=0,5(mol)$
$\overline{M}_{spk}=19,2.2=38,4$
Gọi x, y là số mol $NO$, $N_2O$
$\Rightarrow 30x+44y=38,4(x+y)$
$\Leftrightarrow 8,4x=5,6y$
$\Leftrightarrow \dfrac{x}{y}=\dfrac{2}{3}$
Đặt $n_{NO}=2x; n_{N_2O}=3x$
$Al+4HNO_3\to Al(NO_3)_3+NO+2H_2O$
$8Al+30HNO_3\to 8Al(NO_3)_3+3N_2O+15H_2O$
$\Rightarrow 2x+\dfrac{8}{3}.3x=0,5$
$\Leftrightarrow x=0,05$
$\to n_{NO}=0,05; n_{N_2O}=0,15(mol)$
b,
$n_{Al(NO_3)_3}=n_{Al}=0,5(mol)$
$m_{dd\text{spứ}}=13,5+200-0,05.30-0,15.44=205,4g$
$\to C\%_{Al(NO_3)_3}=\dfrac{0,5.213.100}{205,4}=51,85\%$