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22 June, 21:25

A sample is a mixture of kcl and kbr. when 1.169 g of the sample is dissolved in water and reacted with excess silver nitrate, 1.892 g of solid is obtained. what is the composition by mass percent of the mixture?

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  1. 23 June, 00:18
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    Let x be the amount in grams of KCl and y be the amount of KBr. The reaction of each with Silver Nitrate is:

    KCl + AgNO₃ → AgCl + KNO₃

    KBr + AgNO₃ → AgBr + KNO₃

    The solid referred to here are AgCl and AgBr. The solution is as follows:

    x+y = 1.169 g or y = 1.169 - x - - > eqn 1

    amount of AgCl = x (1 mol/74.55 g KCl) (1 mol AgCl/1 mol KCl) (143.32 g AgCl/mol) = 1.9225x

    amount of AgBr = y (1 mol/119 g KBr) (1 mol AgBr/1 mol KBr) (187.77 g AgBr/mol) = 1.5779y

    Thus,

    1.9225x + 1.5779y = 1.892 - - > eqn 2

    Substituting eqn 1 to eqn 2,

    1.9225x + 1.5779 (1.169 - x) = 1.892

    Solving for x,

    x = 0.1377 g KCl

    y = 1.169 - 0.1377 = 1.0313 g KBr

    Mass Percent of KCl = 0.1377 / (0.1377+1.0313) * 100 = 11.78% KCl

    Mass Percent of KBr = 1.0313 / (0.1377+1.0313) * 100 = 88.22% KBr
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