A 0.3146-g sample of a mixture of nacl (s) and kbr (s) was dissolved in water. the resulting solution required 40.30 ml of 0.08765 m agno3 (aq) to precipitate the cl - (aq) and br - (aq) as agcl (s) and agbr (s). calculate the mass percentage of nacl (s) in the mixture.

The chemical reaction involved are expressed as:

NaCl + AgNO3 = NaNO3 + AgCl

KBr + AgNO3 = KNO3 + AgBr

The amount of AgNO3 that reacted with the solution reacts with the two components in the solution thus the total amount is the sum of the amount reacted with NaCl (x) and amount reacted with KBr (y)

0.08765 M AgNO3 (.04030 L) = 3.5323 x10^-3 mol AgNO3

3.5323 x10^-3 = x + y

We have two unknowns so we need to set up one more equation to solve this problem.

0.3146 g = mass NaCl + mass KBr

0.3146 = x (1 mol NaCl / 1 mol AgNO3) (58.44 g / 1 mol) + y (1 mol KBr / 1 mol AgNO3) (119 g / 1 mol)

0.3146 = 58.44x + 119y

Solving for x and y,

x = 1.7461 x10^-3 mol AgNO3 reacts with NaCl

y = 1.7862 x10^-3 mol AgNO3 reacts with KBr

mass of NaCl = 58.44x = 0.1020 g

% NaCl = 0.1020 / 0.3146 x 100 = 32.44%