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**Stoichiometry problem**

## Homework Statement

1 g of H

_{2}O

_{2}solution containing x % of H

_{2}O

_{2}by weight required x ml of KMnO

_{4}for complete oxidation in acid medium. What is the normality of KMnO

_{4}solution?

## Homework Equations

H

_{2}O

_{2}+ KMnO

_{4}=> 2KOH + 2MnO

_{2}+ 2O

_{2}

## The Attempt at a Solution

1 g of H

_{2}O

_{2}solution contains x% of H

_{2}O

_{2}.

Molar mass of H

_{2}O

_{2}= 34 g, which means in 100 g of solution, 34g of H

_{2}O

_{2}are present. In 1 g of solution, 0.34 grams of H

_{2}O

_{2}are present. This reacts with 34 ml of KMnO

_{4}.

I know that 17 g of H

_{2}O

_{2}will react with 158.03 g of KMnO

_{4}, 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.

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