Question

A 0.8612 g sample of a mixture consisting of NaBr, NaI, and NaNO3 was analyzed by adding AgNO3 to precipitate the Br¡V and I¡V, yielding a 1.0186 g mixture of AgBr and AgI. The precipitate was then heated in a stream of Cl2, converting it to 0.7125 g of AgCl. Calculate the % (w/w) NaNO3 in the sample

Answer 1

Two equations in two unknowns. Solve simultaneously for X and Y.
Let X = mass NaBr
and Y = mass NaI
mm = molar mass
am = atomic mass

eqn 1:
X(mm AgBr/mm NaBr) + Y(mm AgI/NaI) = 1.0186

eqn 2:
X(2*mm AgCl/2*mm NaBr) + Y(2*mm AgCl/2* mm NaI) = 0.7125

Solve for X = NaBr and Y = NaI. Add these two masses, subtract from the sample mass which gives you the mass of NaNO3.
Then %NaNO3 = (g NaNO3/mass sample)*100 = ?
Post your work if you get stuck.

Answer 2

I am stuck on step 3. I don't know how to calculate.
I am assume there are 3 steps in calculates.

1. x+y+z = 0.8612
2. 0.009191x+ 0.0066716y= 1.0186
3. X(2*mm AgCl/2*mm NaBr) + Y(2*mm AgCl/2* mm NaI) = 0.7125

Thanks!