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chemistry

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If 267 mL of O2, measured at STP, are obtained by the decomposition of the KClO3 in a 3.33 g mixture of KCl and KClO3, 2KClO3(s) --> 2KCl(s) + 3O2

What is the percent by mass of KClO3 in the mixture?

  • chemistry - ,

    Given that 1 mole of any gas occupies 22.4 litres at STP you can calculate the number of moles of O2 in 267 ml, (F moles)

    2KClO3(s) --> 2KCl(s) + 3O2

    tells you that 3 moles of O2 came from 2 moles of KClO3, therefore you can work out how many moles of KClO3 would be needed to yield F moles of O2

    number of moles of KClO3
    =Fx2/3

    Calcluate the molar mass of KClO3, M.

    Thus the mass of KClO3 needed is

    (Fx2xM/3) g

    Hence the percentage of KClO3 in the mixture is

    Fx2xMx100%/(3x3.33)

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