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16 February, 02:46

A chemist weighed out 5.14 g of a mixture containing unknown amounts of bao (s) and cao (s) and placed the sample in a 1.50-l flask containing co2 (g) at 30.0 oc and 750. torr. after the reaction to form baco3 (s) and caco3 (s) was completed, the pressure of co2 (g) remaining was 230. torr. calculate the mass percentages of cao (s) and bao (s) in the mixture. 86.6 % bao; 13.4 % cao

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  1. 16 February, 05:42
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    Let the mass of CaO = x grams

    So mass of BaO = 5.14 - x grams

    moles of CaO = mass / molar mass = x / 56

    Moles of BaO = mass / molar mass = 5.14-x / 153

    initial moles of CO2 = PV / RT = 750 X 1.50 / 760 X 0.0821 X 303 = 0.06

    final mole sof CO2 = PV / RT = 230 X 1.50 / 760 X 0.0821 X 303 = 0.018

    So moles of BaCO3 and CaCO3 formed = 0.06 - 0.018 = 0.042 moles

    x / 56 + (5.14-x) / 153 = 0.042

    on solving

    x = 0.68

    So mass of CaO = 0.68 g

    So percentage of CaO = 0.68 X 100 / 5.14 = 13.4 %

    Percentage of BaO = 86.6%
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