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?

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