How Do You Calculate the Normality of KMnO4 in This Stoichiometry Problem?

[Quantum Mind]

Stoichiometry problem

Homework Statement

1 g of H₂O₂ solution containing x % of H₂O₂ by weight required x ml of KMnO₄ for complete oxidation in acid medium. What is the normality of KMnO₄ solution?

Homework Equations

H₂O₂ + KMnO₄ => 2KOH + 2MnO₂ + 2O₂

The Attempt at a Solution

1 g of H₂O₂ solution contains x% of H₂O₂.

Molar mass of H₂O₂ = 34 g, which means in 100 g of solution, 34g of H₂O₂ are present. In 1 g of solution, 0.34 grams of H₂O₂ are present. This reacts with 34 ml of KMnO₄.

I know that 17 g of H₂O₂ will react with 158.03 g of KMnO₄, based on law of fixed proportions. The problem here is only the volume has been given and not the weight. The formula for Normality i.e. 1 mol / 1 L whereby 34 ml or 0.034 L is substituted in the denominator is not correct.

I know I am doing something wrong, but can't figure out what it is. Any help is appreciated.

The answer is 0.58 N.

I am struggling with stoichiometry and I hope that I will be allowed to post thorny (from my point of view) problems in this thread.

 

[Borek]

First of all - your reaction equation is wrong. Permanganate in acidic solution gets reduced to Mn²⁺.

Molar mass of H₂O₂ = 34 g, which means in 100 g of solution, 34g of H₂O₂ are present.

No, 1g of the solution contains 0.x g of H₂O₂

I am struggling with stoichiometry and I hope that I will be allowed to post thorny (from my point of view) problems in this thread.

Please start a new thread for each problem.

 

[Quantum Mind]

First of all - your reaction equation is wrong. Permanganate in acidic solution gets reduced to Mn²⁺.

Can you give the equation?

No, 1g of the solution contains 0.x g of H₂O₂

I am beginning to understand. Guess my whole understanding was off. Since it says that the solution contains x % of hydrogen peroxide, I can't assume it as 34% by weight. Is this correct?

The solution is to take the weight of hydrogen peroxide as x/100 in 1 gram. Since the equivalent weight of H2O2 is 17, the number of equivalence in this solution would be x/1700.

How do I proceed beyond this?

 

[Borek]

Can you give the equation?

Come on, you can. And if you don't know how to balance the equation - its time to learn it. Products are Mn²⁺, H₂O and oxygen. Potassium is just a spectator, if you will balance net ionic equation it will not interfere. And don't forget solution is highly acidic, you will need a lot of acid on the left.

I am beginning to understand. Guess my whole understanding was off. Since it says that the solution contains x % of hydrogen peroxide, I can't assume it as 34% by weight. Is this correct?

The solution is to take the weight of hydrogen peroxide as x/100 in 1 gram. Since the equivalent weight of H2O2 is 17, the number of equivalence in this solution would be x/1700.

2xYes.

How do I proceed beyond this?

Express number of equivalents in x mL of the titrant solution as a function of x and concentration. You will have two equations in two unknowns.

 

[Quantum Mind]

Equivalents in KMnO4 will be Nx/1000

x/1700 = Nx/1000

N = 10/17 or 0.58 N.

Got it, thanks.