Consider the reaction represented by the following equation: xKMnO 4(aq) + ySO2(g) + zH2O(I) → K 2SO4(AQ) + 2MnSO4(aq) + 2H2SO x,y and z are respectively

The equation is garbled in the question, so I first reconstruct it. The standard reaction is:

KMnO₄ + SO₂ + H₂O → K₂SO₄ + MnSO₄ + H₂SO₄

The "2" before MnSO₄ and H₂SO₄ in the question is most likely a typing error. The balanced equation is:

2KMnO₄ + 5SO₂ + 2H₂O → K₂SO₄ + 2MnSO₄ + 2H₂SO₄

Step 1: Use redox (electron) balancing

• Mn goes from +7 in KMnO₄ to +2 in MnSO₄. It gains 5 electrons.
• S goes from +4 in SO₂ to +6 in sulfate. It loses 2 electrons.
• The lowest common multiple of 5 and 2 is 10. So we need 2 Mn and 5 S.
• This gives x = 2 and y = 5.

Step 2: Balance K

• 2KMnO₄ gives 2 K, so we need 1 K₂SO₄.

Step 3: Balance S

• Left side: 5 S from SO₂.
• Right side: 1 (K₂SO₄) + 2 (MnSO₄) + 2 (H₂SO₄) = 5 S. ✔

Step 4: Balance H and O (find z)

• H: 2H₂SO₄ has 4 H, so we need 2H₂O on the left. This gives z = 2.
• O on the left: 2×4 + 5×2 + 2 = 20.
• O on the right: 4 + 8 + 8 = 20. ✔

So x = 2, y = 5, z = 2.

Tip: In redox balancing, cross over the electron changes. Mn changes by 5 and S changes by 2, so Mn gets the coefficient 2 and S gets the coefficient 5. Then balance K, S, H and O in that order, checking oxygen last.