Balance the redox reaction by oxidation number or ion electron method and indicate oxidizing and reducing agent . KMno4 + KBr + H2SO4 approach K2SO4 + MnSO4 + H2O + Br2​

2KMnO₄ (aq) + 10KBr (aq) + 8H₂SO₄ (aq) -> 2MnSO₄ (aq) + 5Br₂ (aq) + 6K₂SO₄ (aq) + 8H₂O (l)

oxidizing agent = KMnO₄

reducing agent = KBr

Explanation:

KMnO₄ -> 2Mn⁷⁺ + 10e⁻ = 2Mn²⁺

KBr -> 10Br⁻⁺ - 10e⁻ = 10Br⁰

3 mol

Explanation:

Given data:

Moles of hydrogen = 3 mol

Moles of oxygen = 2 mol

Yield of reaction = 80%

Moles of water produced = ?

Solution:

Chemical equation:

2H₂ + O₂  →  2H₂O

Now we will compare the moles of water with both reactants.

                       H₂           :         H₂O

                        2            :            2

                          3          :             3

                          O₂        :           H₂O

                           1          :              2

                           2          :         2×2 =   4

Hydrogen produced less number of moles of water so it will limiting reactant.

So number of moles of water produced are = 3 mol