 M O2 = 3,33 – 2,13 = 1,2  N O2 = 1,2/32 = 0,0375 MOL  N HCL = 2...

Câu 15:  m O2 = 3,33 – 2,13 = 1,2  n O2 = 1,2/32 = 0,0375 mol  n

HCl

= 2x + 2y + 3z = 0,15 mol  V

HCl

= 0,15/2 = 0,075 lít = 75 ml Nh n xét : n

O

= n

H2O

; n

HCl

= 2. n

H2O

 n

HCl

= 4.n

O2

= 4.0,0375= 0,15  V = 75 ml