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    Hi,

    I have a problem with this question.

    Hydrogen peroxide readily decomposes to give water and oxygen.
    Hydrogen peroide is sold by volume strength. For example 20-Volume H202 (hydrogen peroxide) yields 20 volumes of oxygen gas for each volume H2O2 solution.

    i.) construct an equation for the decomposition of Hydrogen peroxide
    I can do this part
    ................................ ................................ ................

    ii.) Determine the concentration in mol dm-3, of 20 volume Hydrogen peroxide. Show all your working clearly.

    Any one help casue i have no idea where to start.
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    (Original post by Redeyejedi)
    Hi,

    I have a problem with this question.

    Hydrogen peroxide readily decomposes to give water and oxygen.
    Hydrogen peroide is sold by volume strength. For example 20-Volume H202 (hydrogen peroxide) yields 20 volumes of oxygen gas for each volume H2O2 solution.

    i.) construct an equation for the decomposition of Hydrogen peroxide
    I can do this part
    ................................ ................................ ................

    ii.) Determine the concentration in mol dm-3, of 20 volume Hydrogen peroxide. Show all your working clearly.

    Any one help casue i have no idea where to start.
    you didn't state unit for volume i.e cubic centimeter or litres?
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    dm^3

    oh and if you want to see the question its here..

    http://www.emailed.org.uk/exams/2816_1_Jan_04QP.pdf

    questions 2c and 2d

    thanks
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    i assume that by 20 volumes of O2 for each volume of H2O2 the question means that volume ratio of O22O2 is 20:1
    therefore say that means 1dm³ H2O2 and 20 dm³ O2

    1mole of O2 occupies 24dm³
    therefore
    20/24=0.833 moles O2

    as the mole ratio O22O2 is 1:2 (1mole of O2 is produced from every 2moles of H202
    there are 0.833*2=1.67moles H2O2

    Concentration = moles/volume
    =1.67/1
    =1.67moldm-3

    i'm not at all sure whether thats right but it's the best i can come up with
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    Thanks for your help
 
 
 
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