[Rutherford]

First, thanks for helping me to solve my chemical problems. Now I have 2 more (very hard for me):
1.A mixture of 2 pottasium halides has mass 7,63g. It reacts with AgNO3 ion water and 5,64g of precipitate is made. There were also made 0,1 mol of potassium ions. What salts were in the mixture and what was their mass?
2.A platinum plate is coated with a divalent metal. It reacts with a Bi(NO3)3 solution (Bi is separated on the plate) and the increase in mass of the plate was 2,083g. Then the plate is immersed in a AgNO3 solution (Ag is separated on the plate) and the mass increase was 1,917g now. Calculate the Ar of the divalent metal.

[Arkcon]

You've been on these forums a while, and you've been told at least once, we don't give complete answers on demand.  Writing two, unrelated questions, without any work at all is really flouting the Forum Rules

To begin to solve number one, can you start to relate the text to the numbers?  For example, the first sentence mentions the mass of one reactants, the last sentence mentions the moles of one product, the sentences are split up by other information.  You should try to quickly organize your word problems on paper so you can solve them -- this is a common examination trick.  You will encounter it in many courses.

[Rutherford]

I don't expect a complete answer, just some guidelines. I am not sending the 1st problem I encounter. I was trying to solve those two many times and I couldn't. I tried the 1st one this way:
aKX,bKY
a*(39+x)+b*(39+y)=7.63g
a+b=0,1mol
a*(108+x)+b*(108+y)=5.64g
3 equations and 4 unkowns, I couldn't find another equation.
and the 2nd one:
3M+2Bi³⁺-->3M²⁺+2Bi 2.083g
Bi+3Ag⁺-->Bi³⁺+3Ag 1.917g
1: 2*209y-3x=2.083
2: 3*108z-209y=1,917
And again I have more unknows than equations.

[Borek]

First one needs a trial and error. There is only a limited number of halides, I would start checking each of the three pairs from Cl, Br and I.

[Rutherford]

In the answer the salts are KCl and KBr. I figured out by looking at the answer that the 5,64g of precipitate is only AgCl because AgBr is ionizated in the solution (Ag+,Br-), but how could I know this before looking at the answer?

[Borek]

AgBr is even less soluble than AgCl. And AgI is less soluble than AgBr.

[Rutherford]

Then the precipitate that is made can't have the mass 5,64g, because it is said that o,1mol of K is made so it means that 0,1mol of Ag reacted and if both salts are insoluble then the mass should be bigger than 5,64g.

[Borek]

Well, of number don't fit, then perhaps there EXIST a soluble halide. What halides do you know?

[Rutherford]

Oh, then I had to guess that one of the pottasium halides is KF, so I have 1 unknown less and it can be solved. Thank you very much.

[Hunter2]

I dont think it is flouride, because AgF has good solubility, approx 1,8 kg/l. Only AgCl, AgBr and AgI has bad solubilty.

[Borek]

I dont think it is flouride, because AgF has good solubility, approx 1,8 kg/l.

Who told you precipitate contains all halides?

[Hunter2]

The precipitate cannot contain flouride, because of it good soloubility.

The case could be flouride stays in the solution and only one other halide which precipitates.

[Rutherford]

I have meant that AgF is soluble and now I have to find out what is the insoluble one (5,64g).

[Hunter2]

Ok that makes it more clear now.