my questions are listed before .it for an exam

Question

A 0.6047 g sample containing chloride and iodide ions was treated with silver nitrate giving a 0.4430 g silver halide precipitate. This precipitate was strongly heated in a stream of chlorine gas converting AgI to AgCl; after this process, the precipitate weighed 0.3181 g. Calculate the % Cl and % I in the sample. Hint: This problem requires you to solve a pair of simultaneous equations.  Partial credit will be awarded for setting up word equations and/or sketch of the process, even if you cannot solve algebraically. 

Answer 1

Let W_AgCl and W_AgI be the mass of AgCl and AgI respectively in the silver halide precipitate.

After treating with Cl2 gas, the AgI is converted to AgCl. The mass of AgCl produced by that reaction is given by mol. mass AgCl / mol. mass AgI x W_AgI = 0.6105 x W_AgI. (2AgI + Cl₂ ==> 2AgCl + I₂)

We can set up simultaneous equation:

W_AgCl + W_AgI = 0.4430g

W_AgCl + 0.6105 x W_AgI = 0.3181g

Solving by elimination: W_AgI - 0.6105 W_AgI = 0.1249 g or W_AgI = 0.3207 g

The mass of AgCl, W_AgCl is 0.4430 - 0.3207 = 0.1223 g

The %Cl in the sample is atom. mass Cl / mol. mass AgCl x 0.1223 / 0.6047 x 100

%Cl = 35.45 / 143.3 x 0.1223 /0.6047 x 100 = 5.003%

%I = atom. mass I / mol. mass AgI x 0.3207/ 0.6047 x 100 = 126.9 / 234.8 x 0.3207 / 0.6047 x 100 = 28.66%

Please recheck math as I had to correct myself several times!