[Stephen]

I think I did this exercise with KBr and KCl maybe
KCl+KBr=3.595 let x+y=3.595

If 2KBr+Cl₂=2KCL+Br₂
then:
m(KBr)=3.595-x and agter reaction mas KCl is equal 3.129 grams-x
then appling stechiometry:
(3.595-x):(3.129-x)=119:74.5
44.5x=104.5235
x=2.35 and this is mass of KCl in a mixture so there's 1.245 grams of potassium bromine
3.595 : 100%=1.245: x%
x=34.63%

Ufff

[sjb]

Not

x*4.788=y*3.129...

but 4.788x + 3.128y = 3.595, and x + y = 1; I think.

[Stephen]

Not

x*4.788=y*3.129...

but 4.788x + 3.128y = 3.595, and x + y = 1; I think.

I don't think so 

[Stephen]

Not

x*4.788=y*3.129...

but 4.788x + 3.128y = 3.595, and x + y = 1; I think.

I calculated x and y using this but not same as I calculated by my way :S
x=28.14% and y=71.86% :S
I wonder what is correct ?

[Borek]

A mixture of two halides of potassium has mass equal 7.63 grams and that mixture was first solubled in a water and than teated with excess of AgNO3(aq).There was 5.64 grams of participate.In filtrate there was 0.1mol of K+ ions.What salts were in a mixture and what are their weights?

Looks impossible - mass of precipitate should be higher than mass of potassium halide, after all you are replacing K⁺ with Ag⁺. There were 0.1 mole of K⁺ in the original mixture, so there should be 0.1 mole of Ag⁺ in the precipitate - that means it should weight 10.79 g WITHOUT halides. 0.1 mole of K⁺ has a mass of 3.91 g, so there were 7.63-3.91=3.72 g halides in the original sample and the precipitate mass was 3.72+10.79=14.51 g.

There must be an error in the original question (I have seen it and I was able to understand enough to know what you have posted is a correct rendering of the original).

[Dan]

There were 0.1 mole of K⁺ in the original mixture, so there should be 0.1 mole of Ag⁺ in the precipitate - that means it should weight 10.79 g WITHOUT halides.

Not necessarily, AgF is water soluble - which is the key to the question...

[Borek]

Oops... nothing like to make an idiot out of yourself 

[Stephen]

I still don't understand how am I suppose to do this !?

[Dan]

I've been looking at this and it is a bit fiddly - there is probably an easier way to do it, but here's what I did:

As Borek said, since we have 0.1 mol of potassium, we must have 3.91 g of potassium.

That leaves us with 7.63-3.91 = 3.72 g of halide.

We know that one of the halides must be F^- because, as Borek also pointed out, the mass of the precipitate is too low to contain 0.1 mol of silver, so one of the silver halides must be soluble - AgF is the only soluble silver halide.

So, where m_A = mass of element A, X is the unknown halogen and M_r is relative mass of unknown halogen:

m_F + m_X = 3.72 (equation 1)

We also know that if we have 0.1 mol of potassium, we must have 0.1 mol of halides:

m_F/19 + m_X/M_r = 0.1 (equation 2)

Considering the precipitate 5.64 g of AgX:

m_X + m_Ag = 5.64 (equation 3)

Since the moles of Ag = moles of X:

m_Ag/108 = m_X/M_r
or    m_Ag = 108m_X/M_r (equation 4)

Substitute equation 4 into equation 3 and we get:

m_X + (108m_X/M_r) = 5.64

Which rearranges to:

M_r = 108m_X/(5.64-m_X) (equation 5)

Substitute equation 5 into equation 2 and we get:

m_F/19 + [(5.64 - m_X)/108] = 0.1

Which rearranges to:

5.68m_F - m_X = 5.16 (equation 6)

Equations 6 and 1 can be solved (simultaneous equations) to find the values of m_F and m_X.

Once you know these values you can use equation 2 to determine M_r - compare this value to your periodic table to identify the halogen.

Like I say, this is really long winded and there is probably an easier more obvious approach to this problem that I've missed.

[cliverlong]

I've been looking at this and it is a bit fiddly - there is probably an easier way to do it, but here's what I did:

As Borek said, since we have 0.1 mol of potassium, we must have 3.91 g of potassium.

That leaves us with 7.63-3.91 = 3.72 g of halide.

We know that one of the halides must be F^- because, as Borek also pointed out, the mass of the precipitate is too low to contain 0.1 mol of silver, so one of the silver halides must be soluble - AgF is the only soluble silver halide.

<< snip details >>

Well spotted. I didn't get to the point establishing AgF must be one of the products. Subtle (sneaky even).

Clive

[Stephen]

I've been looking at this and it is a bit fiddly - there is probably an easier way to do it, but here's what I did:

As Borek said, since we have 0.1 mol of potassium, we must have 3.91 g of potassium.

That leaves us with 7.63-3.91 = 3.72 g of halide.

We know that one of the halides must be F^- because, as Borek also pointed out, the mass of the precipitate is too low to contain 0.1 mol of silver, so one of the silver halides must be soluble - AgF is the only soluble silver halide.

So, where m_A = mass of element A, X is the unknown halogen and M_r is relative mass of unknown halogen:

m_F + m_X = 3.72 (equation 1)

We also know that if we have 0.1 mol of potassium, we must have 0.1 mol of halides:

m_F/19 + m_X/M_r = 0.1 (equation 2)

Considering the precipitate 5.64 g of AgX:

m_X + m_Ag = 5.64 (equation 3)

Since the moles of Ag = moles of X:

m_Ag/108 = m_X/M_r
or    m_Ag = 108m_X/M_r (equation 4)

Substitute equation 4 into equation 3 and we get:

m_X + (108m_X/M_r) = 5.64

Which rearranges to:

M_r = 108m_X/(5.64-m_X) (equation 5)

Substitute equation 5 into equation 2 and we get:

m_F/19 + [(5.64 - m_X)/108] = 0.1

Which rearranges to:

5.68m_F - m_X = 5.16 (equation 6)

Equations 6 and 1 can be solved (simultaneous equations) to find the values of m_F and m_X.

Once you know these values you can use equation 2 to determine M_r - compare this value to your periodic table to identify the halogen.

Like I say, this is really long winded and there is probably an easier more obvious approach to this problem that I've missed.

I can just say WoW and one big thank you, not just for you, for all who were helping me to solve this problem...
I calcualte and it's correct Mr=79.59 and that's Br...
Just to tell you that I don't think there's easier way for solving because this is test for republic competition and that's uper uper uper knowlagde test So it looks on our chemist to give us some exercise as this is .
Big thank for everyone