Hòa tan hoàn toàn hh X gồm 8,1 gam Al và 29,25 gam Zn bằng dd HNO3 12,6%

$m_{Al{\left(N{O}_3\right)}_3} = \frac{8,1.213}{27} = 63,9 gam; m_{Zn{\left(N{O}_3\right)}_3} = \frac{29,25.189}{65}$ = 85,05 gam

m_y = $m_{Al{\left(N{O}_3\right)}_3}+ m_{Zn{\left(N{O}_3\right)}_3} + m_{N{H}_{4}N{O}_3} \Rightarrow m_{N{H}_{4}N{O}_3} = 6 gam \Rightarrow n_{N{H}_{4}N{O}_3 }$= 0,075 mol

$n_{N_{2}O} = \frac{(0,3.3+0,45.2-0,075.8)}{8}=$0,15 mol ⇒ V = 0,15.22,4 = 3,36 lít