36.5k views
2 votes
4. 8.492g of ammonium iron (II) sulphate, (NH4)2SO4.FeSO4.nH2O were dissolved

and made up to 250cm³ of solution. A 25.0cm³ portion was acidified and titrated
against 0.0150 moldm3 potassium manganate (VII) solution. A volume of
22.5cm³ was required.
Determine the value of n.

4. 8.492g of ammonium iron (II) sulphate, (NH4)2SO4.FeSO4.nH2O were dissolved and-example-1

1 Answer

5 votes

The value of n for the chemical reaction is 13.67.

How to find values?

The number of moles of potassium manganate (VII) used:

moles of KMnO₄ = volume of KMnO₄ × molarity of KMnO₄

moles of KMnO₄ = 22.5 cm³ × 0.0150 moldm³

= 0.0003375 mol

The balanced chemical equation for the reaction between ammonium iron (II) sulphate and potassium manganate (VII) is:

5FeSO₄ + 2KMnO₄ + 8H₂SO₄ → K₂SO₄ + 2MnSO₄ + Fe₂(SO₄)₃ + 8H₂O

From the balanced equation, 5 moles of FeSO₄ react with 2 moles of KMnO₄. Therefore, the number of moles of FeSO₄ in the solution is:

moles of FeSO₄ = (moles of KMnO₄ × 5) / 2

moles of FeSO₄ = (0.0003375 mol × 5) / 2

= 0.00084375 mol

The molarity of FeSO₄ in the solution is:

molarity of FeSO₄ = moles of FeSO₄ / (volume of solution / 1000)

molarity of FeSO₄ = 0.00084375 mol / (250 cm³ / 1000)

= 0.003375 mol/dm³

The molar mass of FeSO₄.nH₂O is:

molar mass of FeSO₄.nH₂O = mass of FeSO₄.nH₂O / (molarity of FeSO₄ × volume of solution / 1000)

molar mass of FeSO₄.nH₂O = 8.492 g / (0.003375 mol/dm³ × 250 cm³ / 1000)

= 396.0 g/mol

The molar mass of FeSO₄ is 152 g/mol. Therefore, the value of n is:

n = (molar mass of FeSO₄.nH₂O - 152 g/mol) / 18 g/mol

n = (396.0 g/mol - 152 g/mol) / 18 g/mol

= 13.67

Therefore, the value of n is 13.67.

User Shirish Coolkarni
by
7.3k points