m = 13,3 g amestec NaCl + KCl
m.pp. = 28,7 g AgCl
a g b g
NaCl + AgNO3 --> NaNO3 + AgCl
58,5 143,5
c g d g
KCl + AgNO3 --> KNO3 + AgCl
74,5 143,5
=>
a + c = 13,3 => c = 13,3 - a
b + d = 28,7
=>
b = 2,45a ; d = 1,93c
=>
2,45a + 1,93(13,3-a) = 28,7
=> a = 3,031/0,52 = 5,83 g NaCl
=> b = 2,45x5,83 = 14,28 g
=> c = 13,3-5,83 = 7,47 g KCl