In an experiment, 4.9g of copper oxide news obtained nfrom 3.92g  of copper . In another experiment,4.55g of copper oxide gave, on reduction, 3.64 g of copper . Show that these figure verify the law of constant proportion

Posted by Tanveer Singh (Jan 12, 2017 2:42 p.m.) (Question ID: 1310)

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  • Answers:
  • Case I

    Mass of copper= 3.92g

    Mass of copper oxide= 4.9g

    % of copper in oxide = (Mass of copper/Mass of copper oxide) x 100

    = 80%

    So, % of oxygen = 100-80= 20%

     

    Case II

    Mass of copper= 3.64g

    Mass of copper oxide= 4.55g

    % of copper in oxide = (Mass of copper/Mass of copper oxide) x 100

    = 80%

    % of oxygen= 100-80=20%

    Thus,copper oxide prepared by any of the given methods contains copper and oxygen in the same proportion by mass.Hence, it proves the law of constant proportions.

    Answered by Shweta Gulati (Jan 17, 2017 12:19 a.m.)
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