Question

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

Answer 1

Three equations and three unknowns but the way they are set up makes it as if it were two equations and two unknowns.
Let X = mass NaX
and Y = mass NaI
and Z = mass NaNO3
(mm stands for molar mass)
---------------------------
eqn 1 is Z + Y + Z = 0.8612g
eqn 2 is mass AgBr from NaBr + mass AgI from NaI = mass AgBr + mass AgCl. This must be set up in terms of X and Y which is
eqn 2
X(mmAgBr/mmNaBr) + Y(mmAgI/mmNaI) = 1.0186g

eqn 3 is mass AgBr converted to AgCl + mass AgI converted to AgCl = mass AgCl and that set up in terms of X and Y is
eqn 3
X(mmAgCl/mmNaBr) | Y(mmAgCl/mmNaI) = 0.7125g

Solve 2 and 3 simultaneously to obtain X and Y, then plug X and Y into equation 1 to obtain Z (in grams).
Then %Z = (mass Z/mass sample)*100 = ?

Post your work if you have trouble.

Answer 2

29.42%

Answer 3

30.72%

Answer 4

30.72%

Answer 5

Eq2 and 3 can not be solved by simultaneously eq