3X+4nHNO3=3X(NO3)n+nNo+2H2O
10X+12nHNO3=10X(NO3)n+nN2+6nH2o
goi nNO =x,nN2=y
nkhi=5,6:22,4=0,25mol
=>x+y=0,25 *
ta co m khi =30x+28y=7,2**
tu *va **.
=>x=0,1 ,y=0,15
theo pt 1 nHNO3=4nNo=0,4mol
theo pt 2 nHNO3=12nN2=1,8 mol
ta co nHno3=5.0,5=2,5 mol
vi 0,4+1,8<2,5 => Hno3 du X tg het
theo pt 1 nX=0,3/n
theo pt 2 nX=1,5/n
=>nX=1,8/n
=> 1.8/n.MX=16,2
=>X=9n
thu 1<=n<=3
=>X :Al