k2cr2o7 hcl

Writing the oxidation numbers of all the atoms.
+1K2+6Cr2−2O7++1H−1Cl→+1K−1Cl++3Cr−1Cl3++1H2−2O+0Cl2
The Ox. no. of Cr has decreased while that of chlorine has increased.
K2+6Cr2O7→+32CrCl3         .........(i)
H−1Cl→0Cl          ...........(ii)
Decrease in Ox. no. of Cr=6 units per molecule K2Cr2O7
Increase in Ox. no. of Cl=1 unit per molecule HCl
Eq. (ii) is multiplied by 6.
K2Cr2O7+6HCl→2CrCl3+3Cl2
To balance chlorine and potassium, 14 molecules of HCl are required.
K2Cr2O7+14HCl→2CrCl3+3Cl2+2KCl
To balance hydrogen and oxygen, 7H2O are added to tướng RHS.
Hence, the balanced equation is,
K2Cr2O7+14HCl→2KCl+2CrCl3+3Cl2+7H2O.