.

$a)$ Đặt $n_{NO}=x(mol);n_{N_2O}=y(mol)$

$\Rightarrow \dfrac{28x+44y}{x+y}=19,2.2=38,4$

$\Rightarrow 28x+44y=38,4x+38,4y$

$\Rightarrow 10,4x=5,6y$

$\Rightarrow x=\dfrac{7}{13}y(1)$

$\Rightarrow \%V_{NO}=\dfrac{x}{x+y}.100\%=\dfrac{\dfrac{7}{13}y}{\dfrac{7}{13}y+y}.100\%=35\%$

$\Rightarrow \%V_{N_2O}=100-35=65\%$

$b)n_{Al}=\dfrac{13,5}{27}=0,5(mol)$

$Al+4HNO_3\to Al(NO_3)_3+NO\uparrow+2H_2O$

$8Al+30HNO_3\to 8Al(NO_3)_3+3N_2O\uparrow+15H_2O$

$\Rightarrow x+\dfrac{8}{3}y=0,5(2)$

Từ $(1)(2)\Rightarrow x=0,084(mol);y=0,156(mol)$

$\Rightarrow n_{NO}=0,084(mol);n_{N_2O}=0,156(mol)$

Theo PT: $\Sigma n_{HNO_3}=4x+10y=1,896(mol)$

$\Rightarrow C_{M_{HNO_3}}=\dfrac{1,896}{2,2}\approx 0,86M$