Question

What volume of o2(g) measured at 1 atm and 273 k will be formed by action of 100 ml of 0.5 n kmno4 on hydrogen peroxide in an acid solution?the skeleton equation for the reaction is kmno4 +h2so4 +h202--->k2so4 + mnso4+o2+h2o (1)0.12ml (2)0.028ml (3)0.56 ml (4)1.12 litre?

Answer 1

We need a balanced equation of the reaction which in this case is :

2 KMnO₄ + 3 H₂SO₄ + 3 H₂O₂ — > K₂SO₄ + 2 MnSO₄ + 4 O₂ + 6 H₂O

0.5 n KMnO₄ = 0.1 mol / litre

0.1 moles are in 1000 cm³

How many moles are in 100 cm³

(100 / 1000) × 0.1 = 0.01moles

The mole ratio of KMnO₄ to O₂ is :

1 : 2

Moles of Oxygen evolved is thus :

0.01 × 2 = 0.02moles

Molar volume of gas at stp is :

1 mole = 22400 cm³

0.02 moles =?

0.02 × 22400 = 448 cm³


The volume of oxygen is thus 448 cm³

Answer 2

Answer:

(3) 0.56 ml

Explanation:

100 mL of 0.5 N KMnO4 solution gives 0.5*100mL/1000 equivalents [Normality is equivalents/ litre]

0.05 equivalents of KMnO4 would give 0.05 equivalents of O2

n-factor for O2 is 2 as H2O2 loses 2 electrons

Hence, 0.05/2 = 0.025 moles of O2 will be produced

Molar volume = 22.4 litres

Volume of 0.025 moles of O2 = 0.025*22.4 = 0.56 litres