[Kuahji]

This was a pretty long double titration problem, I solved it all except for the last part & could use a bit of help because I'm stuck.

Part 1 A 100.00 ml solution containing aqueous HCl & HBr was titrated with .1200 M NaOH.  The volume of base required to neutralize the acid was 47.14 mL.  a) Write the net ionic equation for both reactions that occur during the titration.
Answer: 2H+ (aq) + 2(OH)- (aq) -> 2H2O (l)

Part 2 Determine the total number of moles of acid (HBr + HCl) in the original solution.
Answer: .002828 mol HBr & .002828 mol HCl

Part 3 Aqueous AgNO3 was then added to the solution (after the titration).  Write the net ionic equation for both reactions that occur. 
Answer: Br- (aq) + Cl- (aq) + 2Ag+ (aq) -> AgBr (s) + AgCl (s)

Part 4 The total mass of solid produced from the above reactions was .9975 g.  Did the original solution contain more HCl or more HBr (in grams)?
This is the part I'm stuck on.  At first I tried to take the mass percents of both of the solid products, into the total mass of the solid produced.  But when I talked to my professor, she stated there wasn't necessarily equal moles of HCl & HBr in the beginning.  Any ideas?

[Padfoot]

Part 2 Determine the total number of moles of acid (HBr + HCl) in the original solution.
Answer: .002828 mol HBr & .002828 mol HCl

TOTAL amount of acid is OK.

Part 4 The total mass of solid produced from the above reactions was .9975 g.  Did the original solution contain more HCl or more HBr (in grams)?
This is the part I'm stuck on.  At first I tried to take the mass percents of both of the solid products, into the total mass of the solid produced.  But when I talked to my professor, she stated there wasn't necessarily equal moles of HCl & HBr in the beginning.  Any ideas?

The mass of solid produced is 0.9975g.  This is made up of both AgBr and AgCl like you said.
From this info comes the relationship:
n1(M_AgCl)+ n2(M_AgBr)=0.9975

We also know: n1 + n2 = 0.005656 (TOTAL acid)
We know this because n(AgCl) produced = n(HCl) initially
and likewise for n(AgBr).

You should now be able to solve for n1 and n2 which are the amounts of HCl and HBr in your intitial sample respectively.