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
