Question

potassium permanganate + hydrochloric acid => potassium chloride + manganese chloride + chlorine + water balancing ​

Answer 1

Explanation:

This is a classic redox problem involving MnO4- . The key to it is the ionic form of the redox equation for this ion which is:

8 H+ + MnO4- + 5 e- → Mn2+ + 4 H2O

Since we are forming molecular Cl2 , we need another redox ionic equation

2 Cl- → Cl2 + 2 e-

Since the e- must cancel out when we combine these equations, we need to multiply the first one by 2

and multiple the second one by 5 and then combine to get

16 H+ + 2 MnO4- + 10 Cl- → 2 Mn2+ + 8 H2O + 5 CL2

This is the hard part. The rest is just a matter of adding the two K+ ions to each side and six more Cl- ions to each side to take care of the MnCl2 and KCl that will be on the right hand side.

The final result is:

2 K MnO4 + 16 HCl → 2 MnCl2 + 5 Cl2 + 2 KCl + 8 H2O.

PLEASE MARK THE BRAINLIEST

Answer 2

Answer:

2 k Mno4 +16 hcl= 2kcl+2Mncl2+5cl2+8h2o