[Stephen]

Well, there are one type of exercises which I don't even know how to start doing, on first look they sound easy but not for me, These are for competition from chemistry for high school I need to know them...I have Serbian version so don't be angry on me if I don't translate something good.Thank you...

For now I found just two of them on test but I hope I'm not going to need to post more of them after these two
1.It's given mixture of two halogenides of potassium which has weight of  7.63g.The the mixture is dissolved in a water and teated with excessive(an excess) of solution of AgNO3.Then  there appeared 5.64 g of lees(if I can say so ).In filtrate there were 0.1 mol K⁺ ions.What're the name of the salts and what're their masses?

2.Mixture of KBr and KCl has mass of  3.595g and is boiled with chlorine  so all KBr was transformed in KCl.On the end of reaction (when all KBr became KCl), mass of KCl was 3.129g.What was the percentage of KBr in mixture?

So these two are my problems for now...

[cliverlong]

1.1 When you write halogenide do you mean halide?

1.2. If halide, then which elements form halides with potassium? Two of these will be the compounds in your mixture.

1.3 Find out what are the reactions of the various halides with AgNO3 (silver nitrate) . This is a very standard test and there will be loads of examples and explanations of this reaction on the net.

1.4 By lees, do you mean precipitate?

1.5 Do you understand what a filtrate is?

1.6 Are you sure you have given in your post all the information from the original question?


2.1 KBr and KCl are solids. Chlorine is a gas. You cannot "boil" KBr and KCl with chlorine. Do you mean bubble chlorine gas through an aqueoue solution of the mixture?

2.2 Say you are told you have 3.129g of KCl, what other measure can you calculate about the KCl? What mathematical formulae do you know when you are given chemical compounds and masses?

If you can answer the questions above you can make progress with your questions

Clive

[Stephen]

1.1 When you write halogenide do you mean halide?

1.2. If halide, then which elements form halides with potassium? Two of these will be the compounds in your mixture.

1.3 Find out what are the reactions of the various halides with AgNO3 (silver nitrate) . This is a very standard test and there will be loads of examples and explanations of this reaction on the net.

1.4 By lees, do you mean precipitate?

1.5 Do you understand what a filtrate is?

1.6 Are you sure you have given in your post all the information from the original question?


2.1 KBr and KCl are solids. Chlorine is a gas. You cannot "boil" KBr and KCl with chlorine. Do you mean bubble chlorine gas through an aqueoue solution of the mixture?

2.2 Say you are told you have 3.129g of KCl, what other measure can you calculate about the KCl? What mathematical formulae do you know when you are given chemical compounds and masses?

If you can answer the questions above you can make progress with your questions

Clive

I knew I won't translate it well

1.halide, but in our language we call them such as ,,halogenides,, so I thought it might be the same in English
2.I know that's going to be an elements from VII group but just two of them are with potassium in mixture as compound KX where X is element from VII group.
3.If I understand you're looking for this:
KX+AgNO₃--->AgX+KNO₃
4.Yes I do
5.I think I do, unsolvent part of some compound, i think ...
6.Let me try again, but I think I did:


1. A mixture of two halogenides of potassium with a mass of 7,63 grams is dissolved in water and treated with excess AgNO3. 5,63 grams of precipitate. There was 0.1 mols of K+ ions in the filtrate. Which salts were in the mixtures and what are their masses?










2.1 I really don't know right ex preson but my dictionary says teat but I think we can say buble too ...

2.2 I can calculate number of moles, molecules and that's all I guess




2. A mixture of KCl and KBr with a mass of 3,595 grams is heated with chlorine so that the entire KBr reacted into KCl. Total mass of KCl after the reaction is 3,129 grams. Calculate the mass percentage of KBr in the mixture.

[cliverlong]

1.1 When you write halogenide do you mean halide?

1.2. If halide, then which elements form halides with potassium? Two of these will be the compounds in your mixture.

1.3 Find out what are the reactions of the various halides with AgNO3 (silver nitrate) . This is a very standard test and there will be loads of examples and explanations of this reaction on the net.

1.4 By lees, do you mean precipitate?

1.5 Do you understand what a filtrate is?

1.6 Are you sure you have given in your post all the information from the original question?


2.1 KBr and KCl are solids. Chlorine is a gas. You cannot "boil" KBr and KCl with chlorine. Do you mean bubble chlorine gas through an aqueoue solution of the mixture?

2.2 Say you are told you have 3.129g of KCl, what other measure can you calculate about the KCl? What mathematical formulae do you know when you are given chemical compounds and masses?

If you can answer the questions above you can make progress with your questions

Clive

I knew I won't translate it well

1.halide, but in our language we call them such as ,,halogenides,, so I thought it might be the same in English

OK - we know we are talking about the same thing.

2.I know that's going to be an elements from VII group but just two of them are with potassium in mixture as compound KX where X is element from VII group.

Yes, fine
Please list the possible elements of Group 7 - this becomes important later

3.If I understand you're looking for this:
KX+AgNO₃--->AgX+KNO₃

OK - this is central to the question
You need to know something specific about the different compounds AgX when X is any one of the group VII elements.

4.Yes I do
5.I think I do, unsolvent part of some compound, i think ...

Precipitate: is an insoluble solid produced when two liquids are mixed

Suspension: If a solid is produced by mixing two liquids but the particles of the solid remain floating around in the liquid, the mixture is a suspension.

Filtrate: is the liquid left after filtering a suspension of a solid in a liquid.

Residue: is the solid residue remaining in the filter following filtration

6.Let me try again, but I think I did:


1. A mixture of two halogenides of potassium with a mass of 7,63 grams is dissolved in water and treated with excess AgNO3. 5,63 grams of precipitate. There was 0.1 mols of K+ ions in the filtrate. Which salts were in the mixtures and what are their masses?

So, you have KX and KY dissolved.
Mix with AgNO₃(aq)
A precipitate is formed. What does that tell you about AgX or AgY?
Can you write down the two chemical equations that represent the reactions?

2.1 I really don't know right ex preson but my dictionary says teat but I think we can say buble too ...

New dictionary needed?

2.2 I can calculate number of moles, molecules and that's all I guess

Good.
So, how can you apply that ?

[Stephen]

1.F,Cl,Br and I; Not sure is At  important here, but lets say that he can also be one of the halogens.
2.Well, I don't
I'm kiddin' , but I really don't know a lot about all of them, I know that AgCl is insoluble in water and has a white colour, then I know that AgBr and AgJ are also insoluble in a water but they are yellow I think :S And I don't know anything about AgAt And I forgot AgF, I think he is soluble in water!
But something specific, I can't say they are all specific on their own way you have to be more precise so I could know the right way of thinking
3.you helped me a lot with those explanation, thank you !
4.That tells me that those two solids are also insoluble in a water or maybe partly(not at all) soluble in water.
Well I can If I could now X and Y, maybe I'm not informed well but all these solids AgCl,AgBr and AgJ are in very small amount dissolved but just a little.So I think I can say that AgF is not correct answer.And still I don't know anything about AgAt but something's telling me that he is not solution for this problem:S
5.I will buy new one
6.Well I thought something like this:

2KBr+Cl₂=2KCl+Br₂
We see that n(KBr)=n(KCl)
And maybe I can do this:
M=m/n so n(KCl)=3.129/74.6=0,04194 mol so that's also n(KBr)
Now formula up: m(KBr)=4,788 grams and that's when I say ha ha ha ... (irony)

There's something I am missing or not doing correct

[cliverlong]

<< snip >>
 I know that AgCl is insoluble in water and has a white colour, then I know that AgBr and AgJ are also insoluble in a water but they are yellow I think :S And I don't know anything about AgAt And I forgot AgF, I think he is soluble in water!
<< snip >>

Aha !

Well we have two situations

1. A mixture of KF and KX (where X is one of the other halogens)

or

2. A mixture of KX and KY (where X and Y are both halogens but neither is Fluorine)

Now,

What happens  when you add AgNO₃ to situation 1? Write the chemical equation and ensure you include the state symbol (g), (aq), (s) , whatever for every compound.

What happens  when you add AgNO₃ to situation 2? Write the chemical equation and ensure you include the state symbol (g), (aq), (s) , whatever for every compound.

Now, I haven't worked out the detail, as I don't want to give away too much. I will follow you in this problem.
But first a hint: Almost always when performing a chemistry calculation convert everything to moles then use the chemical equation to work out the ratio of moles.

I have a slight suspicion there isn't enough information in your original question to work out this problem - but let's see how far we get.

What data do you have?

7.63g of a mixture of two of KF, KX or KY. Now you don't have enough information here to work out moles of either of the two potassium salts (why?)

You end up with 0.1moles of K⁺ ions in the filtrate. Look at the chemical equations and what can you deduce from this information?

You get 5.64 g of precipitate at the end. What is this compound when you start from situation 1 (KF, KX) and situation 2 (KX, KY) above? I think the problem may only be solvable if one of the original salts is KF (i.e. situation 1).  So I suggest you make the assumption that one of the salts is KF (i.e. situation 1) and see how far you get.

Clive

[cliverlong]

2.Mixture of KBr and KCl has mass of  3.595g and is boiled with chlorine  so all KBr was transformed in KCl.On the end of reaction (when all KBr became KCl), mass of KCl was 3.129g.What was the percentage of KBr in mixture?

A mixture of KBr and KCl weighs 3.595g
The mixture is dissolved in water.
Chlorine gas is bubbled through the solution so all KBr is transformed into KCl.
(What do you see happening in the reaciton vessel?)
The water is evaporated, somehow the bromine is removed and  3.129g of KCl is left.
What was the percentage of KBr in the original mixture?

Hints:
Step 1. Write a chemical equation for the reaction of chlorine with the solution of the two salts. Treat each salt solution as a separate reaction
Step2. What can you wrok out from the information that  3.129g of KCl is produced?
Step3. How can you combine this with the chemical equations?

Clive

[Stephen]

Reactions:
1.KF(aq)+AgNO₃(aq)=KNO₃(aq)+AgF(aq)
KX(aq)+AgNO₃(aq)=KNO₃(aq)+AgX(s)   I guess....But these reactions are parallel, so maybe it's better to write them such as:

(KF+KX)(aq)+AgNO3(aq)=KNO3(aq)+AgF(aq)+AgX(s) Am I suppose to write 2 before AgNO3 and KNO3? :S
2.(KY+KX)(aq)+AgNO3(aq)=KNO3(aq)+AgY(s)+AgX(s) Same question!?

The information 0.1 moles K⁺ says me that I have 0.1moles of KNO₃ and that sum of moles of K+ in mixture is 0.1 moles!?

7.63g of a mixture of two of KF, KX or KY. Now you don't have enough information here to work out moles of either of the two potassium salts (why?)

Hmm Well I don't know what's their ratio in a mixture?!Maybe...


You get 5.64 g of precipitate at the end. What is this compound when you start from situation 1 (KF, KX) and situation 2 (KX, KY) above? I think the problem may only be solvable if one of the original salts is KF (i.e. situation 1).  So I suggest you make the assumption that one of the salts is KF (i.e. situation 1) and see how far you get.

Well in first case it's AgX and in second one it's AgX and AgY!
Well you can suggest me but it's not worth to me because I don't know how to finish exercise:)

I really don't know how to think in this kind of exercise, so I just asked to do for me or help me so I could now just a way of thinkin' in this kind of exercises ....I am staring in this equations but don't see anything useful, I know am wrong but I'm trying to see a solution...
Well you told that maybe it's good idea to suppose that KF its one of component in the mixture!
So, that means that I'm not suppose to do anything with assumption 2...
(KF+KX)(aq)+AgNO3(aq)=KNO3(aq)+AgF(aq)+AgX(s) this is reaction and ...
Hmmm I  think that somehow I can get something useful from mass of perception AgX but again, I repeat, Don't know how to think with these exercises and don't know am I right .

[Stephen]

2.Mixture of KBr and KCl has mass of  3.595g and is boiled with chlorine  so all KBr was transformed in KCl.On the end of reaction (when all KBr became KCl), mass of KCl was 3.129g.What was the percentage of KBr in mixture?

A mixture of KBr and KCl weighs 3.595g
The mixture is dissolved in water.
Chlorine gas is bubbled through the solution so all KBr is transformed into KCl.
(What do you see happening in the reciting vessel?)
The water is evaporated, somehow the bromine is removed and  3.129g of KCl is left.
What was the percentage of KBr in the original mixture?

Hints:
Step 1. Write a chemical equation for the reaction of chlorine with the solution of the two salts. Treat each salt solution as a separate reaction
Step2. What can you work out from the information that  3.129g of KCl is produced?
Step3. How can you combine this with the chemical equations?

Clive

KBr+Cl₂--->KCl+Br₂ and KCl+Cl₂----> no reaction I suppose
So if 3.129 grams of KCl ar formed then, KBr+Cl₂--->KCl+Br₂ this is a key of problem I think...
m(KCl)=3.129 grams=0.042 moles of KCl I also see that n(KBr):n(KCl)=1:1 and that means that we have also 0.042 moles of KBr =4,998 grams But I see where am I wrong but don't know what else should be logical to do, I'm wrong with this 3.129 grams of KCl, here is also part of KCl from beginning and If I could somehow to find KCl on start of reaction I could easily calculate Kbr mass...
But again nothing XD

[cliverlong]

We are given a mixture of two halides of potassium with a weight of  7.63g.
The mixture is dissolved in water and treated with an excess of AgNO₃(aq).
5.64 g of precipitate is formed.
In the filtrate there are 0.1 mol K⁺ ions.

What are the name of the salts and what are their masses?

We are given a mixture of two halides of potassium with a weight of  7.63g.
5.64 g of precipitate is formed.

Situation 1. KF plus KX
In this case the reactions are:
KF(aq) + AgNO₃(aq)  AgF(aq) + KNO₃(aq) (in fact this is a little misleading. All the salts are soluble in water so all ions are free so it is a bit meaningless to talk about any one salt existing in preference to another)
KX(aq) + AgNO₃(aq)  AgX(s) + KNO₃(aq)
If AgX is the precipitate it could be: AgCl, AgBr, AgI (or even AgAt - but Astatine is a rare radioactive element - so let us ignore that possiblity)
Each of these silver halide precipitates has a different formula mass (relative molecular mass). So we cannot determine how many moles of the silver salt are formed
We can't deduce anything from the subtraction 7.63g - 5.64g because we are subtracting the mass of an insoluble silver salt precipitate from the original mass of potassium salts.
We can't deduce anything about the original number of moles from the original mass 7.63g of potassium salts because we don't know the salt KX and we don't know the relative amounts of KF to KX.

In the filtrate there are 0.1 mol K⁺ ions.

So there were 0.1 mol of K⁺ originally - but we don't know how much of this was in the KF and how much was in the KX.

With situation 2: the salts KX and KY (neither X nor Y are Fluorine), we still cannot deduce moles of the original salts since the mass of the precipitate is composed of unknown ratio of AgX and AgY.

So my thought at this stage (I may have missed something) is there is insufficient information in your original question for there to be a unique answer. We could add information, such as "there are equal number of moles of the two potassium salts in the original sample" and try to see if we can solve the problem with a unique solution. If this is not enough then keep adding further information for example
Salts are: KF, KCl
Salts are: KCl, KBr
Salts are: KCl, KI
etc.

If adding this further information means

1. You can calculate a solution
and
2. Each solution for different "additonal information" gives a different answer, then you can be pretty certain the original problem does not have enough information to result in a unique solution
However, if the you calculate the answer for the different pairs of salts and you keep coming up with the same answer , the some of the data you added was unnecessary.

Try it. Also re-read the original question to see if you forgot any data.

Clive

[cliverlong]

A mixture of KBr and KCl weighs 3.595g
The mixture is dissolved in water.
Chlorine gas is bubbled through the solution so all KBr is transformed into KCl.
(What do you see happening in the reaction vessel?)
The water is evaporated, somehow the bromine is removed and  3.129g of KCl is left.
What was the percentage of KBr in the original mixture?

Hints:
Step 1. Write a chemical equation for the reaction of chlorine with the solution of the two salts. Treat each salt solution as a separate reaction
Step2. What can you work out from the information that  3.129g of KCl is produced?
Step3. How can you combine this with the chemical equations?

Clive

KBr(aq) + Cl₂(g)  (you did not write a balanced equation - please retry)

KCl(aq) + Cl₂(g)  KCl(aq) + Cl₂(g) i.e. no reaction (as you wrote)

 3.129g of KCl = 0.042 moles of KCl (I'm assuming you did calculation correctly - can you show your working?)

Think about the mass loss: 3.595g mixture KCl and KBr to 3.129g of KCl . What has happened to this "lost" mass? Think about both reactions. Which chemical accounts for this mass difference? Can you work out the number of moles of this "lost" material? What does that tell you about the number of moles of one of the original salts (KCl or KBr)? Where do you go from there?

Clive

[Stephen]

Uuuh Once again, and if we don'tfind a solution I'll ask my teacher because I don't want to bother you :S
First, this is a test for competition so I guarantee that it has a solution because there is a team of ph D for chemistry who write these exercises so it's checked 100% sure...And I just looked a key for that test they wrote KF 4.06 and KBr 3.57 grams :S Don't know how  don't know why...
Let me write again:

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?


Uhh, I tried to read it on my own language and I read it 5 times, there's no catch I can see...I was translating in a same form it's on my language believe me
If you  understand:
http://helix.chem.bg.ac.rs/mojahemija/srednje/zad2r2003.pdf exercises 7

[sjb]

m(KCl)=3.129 grams=0.042 moles of KCl I also see that n(KBr):n(KCl)=1:1 and that means that we have also 0.042 moles of KBr =4,998 grams But I see where am I wrong but don't know what else should be logical to do, I'm wrong with this 3.129 grams of KCl, here is also part of KCl from beginning and If I could somehow to find KCl on start of reaction I could easily calculate Kbr mass...
But again nothing XD

You're almost there. In the bubbling of Cl₂ through the mixture, all the KBr is converted to KCl, in effect. You end up with KCl, mass 3.129g. I assume that this is 0.042 mol. If you had no KBr in the first place, only KCl, what mass of starting material would you have? If there was no KCl in the first instance, only KBr, what mass would you start with? If you have x % KCl, and y% KBr, what mass would you start with? Can you find another equation to link x and y here?

[Stephen]

A mixture of KBr and KCl weighs 3.595g
The mixture is dissolved in water.
Chlorine gas is bubbled through the solution so all KBr is transformed into KCl.
(What do you see happening in the reaction vessel?)
The water is evaporated, somehow the bromine is removed and  3.129g of KCl is left.
What was the percentage of KBr in the original mixture?

Hints:
Step 1. Write a chemical equation for the reaction of chlorine with the solution of the two salts. Treat each salt solution as a separate reaction
Step2. What can you work out from the information that  3.129g of KCl is produced?
Step3. How can you combine this with the chemical equations?

Clive

KBr(aq) + Cl₂(g)  (you did not write a balanced equation - please retry)

KCl(aq) + Cl₂(g)  KCl(aq) + Cl₂(g) i.e. no reaction (as you wrote)

 3.129g of KCl = 0.042 moles of KCl (I'm assuming you did calculation correctly - can you show your working?)

Think about the mass loss: 3.595g mixture KCl and KBr to 3.129g of KCl . What has happened to this "lost" mass? Think about both reactions. Which chemical accounts for this mass difference? Can you work out the number of moles of this "lost" material? What does that tell you about the number of moles of one of the original salts (KCl or KBr)? Where do you go from there?

Clive

Huh
2KBr+Cl₂=2KCl+Br₂
1 mol KCl has 74.6 g and x moles has 3.129
x=3.129/74.6=0.04194mol KCl
And here comes harder part
Well, there is something you told me after you wrote reactions this aq says me that there is water
Well water disappeared that's logic because an exercises says that mixture is ,,warmed,, with Cl₂ and there's when water disappeared
delta m =3.595-3.129=0.466 grams of water
1 mol of water has 18 grams and x moles of water has 0.466
x=0.466/18=0.258moles of water
And that's where my brain's stop workin'
But there is still something I dont understand:S
Im not sure I'm in right direction :S
I have mixture of two salts for example x and y
and after something x become y' lets say because of different mass but then y+y'=3.129, I think that all this would only be correct if just y=3.129 don't understand really


What does that tell you about the number of moles of one of the original salts (KCl or KBr)? Where do you go from there?

I dont know answer on this

p.s I really like doing an exercises like this you give me a hint I think don't get me wrong that I want of you to do whole exercise for me, I really try to understand this because I want to know principe of doing these exercises, really do...

[Stephen]

m(KCl)=3.129 grams=0.042 moles of KCl I also see that n(KBr):n(KCl)=1:1 and that means that we have also 0.042 moles of KBr =4,998 grams But I see where am I wrong but don't know what else should be logical to do, I'm wrong with this 3.129 grams of KCl, here is also part of KCl from beginning and If I could somehow to find KCl on start of reaction I could easily calculate Kbr mass...
But again nothing XD

You're almost there. In the bubbling of Cl₂ through the mixture, all the KBr is converted to KCl, in effect. You end up with KCl, mass 3.129g. I assume that this is 0.042 mol. If you had no KBr in the first place, only KCl, what mass of starting material would you have? If there was no KCl in the first instance, only KBr, what mass would you start with? If you have x % KCl, and y% KBr, what mass would you start with? Can you find another equation to link x and y here?

KCl+Cl₂=KCl+Cl₂
m(KCl) after no reaction = 3.129=0.042mol if no reaction is accured that means that mass of KCl in non mixture(when there's no KBr) is also equal 3.129g=0.042mol

KBr+Cl₂=KCl+Br₂
m(KCl) after reaction=3.129=0.042 mol that means on the start of reaction I had 0.042 moles of KBr equal 4,788 grams

Hmm x and y very tough question
I will use philosophu a little bit sorry:S
x*4.788=y*3.129
where x+y=1(100%)=>x=100%-y
when I solve this
y=60.47% and x=39.52%
I know this is stupid thinking but that was the first thing I remember :S
And it's not correct as I can see it says that x=34.7% close but not same ((