Asample 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?

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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