2M + 2NHCL → 2MCL N + NH 2 (1) M 2 O N + 2NHCL → 2MCL N + NH 2 O (2)0
1.
0.125
-Ta có PTHH: 2M + 2nHCl → 2MCl n + nH 2 (1)
M 2 O n + 2nHCl → 2MCl n + nH 2 O (2)
0.25
IV.
48
.
4
0 . 2
(2.25)
n
Hmol → n
HCl= 0.4 x 2 = 0.8 mol
22
2
= =
Theo (1) → n
HCl= 0.4 mol →
n
M= 0 n . 4 mol
0
2
Theo (2) → n
HCl= 0.8 – 0.4 = 0.4 mol →
Mn n
On
2