Question

Balance the following equation by oxidation reduction method FeSO4

+
KMnO4+ H2SO4 → Fe2 (SO4)3+ k2SO4+MnSO4+H2O​

Answer 1

`10\, {\rm Fe}\rm SO_4 + 2\, \rm K {Mn} O_4 + 8\, H_2SO_4\\ \to 5\, {Fe} (SO_4)_3 + K_2SO_4 + 2\, {Mn}SO_4 + 8\, H_2O`.

Step-by-step explanation:

Identify the elements with oxidation state changes:

Oxidation states of iron,
`\rm Fe`:

• 
  `+2` in
  `\rm FeSO_4` among the reactants.
• 
  `+3` in
  `\rm Fe_2(SO_4)_3` among the products.
• Change to the oxidation state:
  `+1` (oxidation) for each
  `\rm Fe` atom.

Oxidation state of manganese,
`\rm Mn`:

• 
  `+7` in
  `\rm KMnO_4` among the reactants.
• 
  `+2` in
  `\rm MnSO_4` among the products.
• Change to the oxidation state:
  `(-5)` (reduction) for each
  `\rm Mn` atom.

The change in the oxidation state of
`\rm Mn` is five times the opposite of the change to the oxidation state of
`\rm Fe`. If there are one mole of
`\rm Mn\!` atoms in each mole of this reaction, there would be five times as many
`\rm Fe\!` atoms per mole reaction. In other words:

`\displaystyle 5\, \overset{+2}{\rm Fe}\rm SO_4 + 1\, \rm K \overset{+7}{Mn} O_4 + ?\, H_2SO_4\\ \to (5)/(2)\, \overset{+3}{Fe} (SO_4)_3 + ?\, K_2SO_4 + 1\, \overset{+2}{Mn}SO_4 + ?\, H_2O`.

(Notice that each mole of this reaction would include five times as many
`\rm Fe` atoms as
`\rm Mn` atoms.)

Multiply the coefficients by
`2` to eliminate the fraction:

`\displaystyle 10\, {\rm Fe}\rm SO_4 + 2\, \rm K {Mn} O_4 + ?\, H_2SO_4\\ \to 5\, {Fe} (SO_4)_3 + ?\, K_2SO_4 + 2\, {Mn}SO_4 + ?\, H_2O`.

Find the unknown coefficients using the conservation of atoms.

Reactants:

• 
  `2` potassium
  `\rm K` atoms in two
  `\rm K_2SO_4` formula units.

Therefore, among the products:

• 
  `2` potassium
  `\rm K` atoms in one
  `\rm K_2SO_4` formula unit.

`\displaystyle 10\, {\rm Fe}\rm SO_4 + 2\, \rm K {Mn} O_4 + ?\, H_2SO_4\\ \to 5\, {Fe} (SO_4)_3 + {1}\, K_2SO_4 + 2\, {Mn}SO_4 + ?\, H_2O`.

Products:

• 
  `5 * 3 + 2 + 1 = 18` sulfur
  `\rm S` atoms in five
  `\rm Fe_2(SO_4)_3` formula units, two
  `\rm K_2 SO_4` formula units, and one
  `\rm MnSO_4` formula unit.

Reactants:

• There are already ten
  `\rm S` atoms in that ten
  `\rm Fe(SO_4)_2` formula units. The other
  `18 - 10 = 8` formula units would correspond to eight
  `\rm H_2SO_4` molecules among the reactants of this reaction.

`\displaystyle 10\, {\rm Fe}\rm SO_4 + 2\, \rm K {Mn} O_4 + 8\, H_2SO_4\\ \to 5\, {Fe} (SO_4)_3 + {1}\, K_2SO_4 + 2\, {Mn}SO_4 + ?\, H_2O`.

Products:

• There are
  `8 * 2 = 16` hydrogen
  `\rm H` atoms in that eight
  `\rm H_2SO_4` molecules.

Therefore, among the products:

• There would be
  `16 / 2 = 8` molecules of
  `\rm H_2O`, with two
  `\rm H` atoms in each
  `\rm H_2O\!` molecule.

`\displaystyle 10\, {\rm Fe}\rm SO_4 + 2\, \rm K {Mn} O_4 + 8\, H_2SO_4\\ \to 5\, {Fe} (SO_4)_3 + {1}\, K_2SO_4 + 2\, {Mn}SO_4 + 8\, H_2O`.