Question

One gram of a mixture of potassium and NACL on treatment with excess of silver nitrate gave 2gm of AgCl .what was the composition of the two salts in the original mixture

Answer 1

xKCl + yNaCl + (x+y)AgNO3 = xKNO3 + yNaNO3 + (x+y)AgCl

Let x moles of KCl and y moles of NaCl present in the mixture

The moles of AgCl formed = x+y = 2/143.5 moles

Therefore

x + y = 2/143.5 --------------(1)

Given that, 74.5x +58.5y = 1 -----------(2)

solving the two equations for x,y

multiply first equation by 74.5

74.5x + 74.5y = 2 x 74.5/143.5 = 149/143.5 = 1.038

74.5x + 58.5y = 1

(74.5-58.5)y = 1.038 -1

y = 0.038/16 = 0.00237 moles

mass of NaCl = 0.0237 x 58.5 = 0.139 g = 14%

mass of KCl = 1 - 0.139 = 0.861 g = 86%

The composition of original mixture is 14 % NaCl and 86% KCl

Alternatively,

Let x gm of KCl and y gm of NaCl Present

x + y = 1 ---------------------(1)

KCl + AgNO3 = KNO3 + AgCl

74.5 g of KCl gives 143.5 g of AgCl

x g of KCl would give 143.5x/74.5 = 1.926x g of AgCl

NaCl + AgNO3 = NaNO3 + AgCl

58.5 g of NaCl gives 143.5 g of AgCl

y g of NaCl would give 143.5y/58.5 = 2.453y g of Ag Cl

Given that 2 g of AgCl formed.

1.926x + 2.453y = 2 -------------(2)

Solving equn 1 and 2

1.926x + 1.926y = 1.926 (from equn 1)

(2.453-1.926)y = 2 - 1.926

y = 0.074/0.527 = 0.14 g = 14%

x = 1 - 0.14 = 0.86 g = 86%