1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Stoichiometry problems thread

  1. Jul 5, 2011 #1
    Stoichiometry problem

    1. The problem statement, all variables and given/known data
    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?

    2. Relevant equations

    H2O2 + KMnO4 => 2KOH + 2MnO2 + 2O2

    3. 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.
    Last edited: Jul 5, 2011
  2. jcsd
  3. Jul 5, 2011 #2


    User Avatar

    Staff: Mentor

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

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

    Please start a new thread for each problem.
  4. Jul 5, 2011 #3
    Can you give the equation?

    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?
  5. Jul 6, 2011 #4


    User Avatar

    Staff: Mentor

    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.


    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.
  6. Jul 6, 2011 #5
    Equivalents in KMnO4 will be Nx/1000

    x/1700 = Nx/1000

    N = 10/17 or 0.58 N.

    Got it, thanks.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook