Chemistry Sodium Thiosulphate Calculation Question

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Sugaray
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Can someone help me with this question?

20 litres of air, contaminated with chlorine gas, were bubbled through an excess of aqeous potassium iodide.

Cl(2) + 2KI ---> I(2) + 2KCl

The iodine formed reacted exactly with 45cm(3) of sodium thiosulphate solution (0.100M).

Calculate the volume of chlorine in the sample of air. Assume 1 mole of gas has a volume of 25 litres at room temperature and pressure and that all measurements were taken at r.t.p.

*The answer I get is 2.786 x 10(-6) ..but it's probably wrong! Can anyone confirm? Thanks
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john !!
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what is an equation for sodiom thiosulphate? I'll probably be able to do the calculations, i'm terible with formulae however.
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Sugaray
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2S(2)O(3)²- [aq] + I(2) [aq] ---> 2I- [aq] + S(4)O(6)²- [aq]
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Sugaray
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that's the reaction with iodine anyway I think!
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john !!
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2S(2)O(3)²- [aq] + I(2) [aq] ---> 2I- [aq] + S(4)O(6)²- [aq]

20 litres of air, contaminated with chlorine gas, were bubbled through an excess of aqeous potassium iodide.

Cl(2) + 2KI ---> I(2) + 2KCl

The iodine formed reacted exactly with 45cm(3) of sodium thiosulphate solution (0.100M).

Calculate the volume of chlorine in the sample of air. Assume 1 mole of gas has a volume of 25 litres at room temperature and pressure and that all measurements were taken at r.t.p.

*The answer I get is 2.786 x 10(-6) ..but it's probably wrong! Can anyone confirm? Thanks

---+---

ok.sodium thiosulphate: S4O6 so
45cm^3 = 0.045dm^3
conc. = 0.1M
so there are 0.045*0.1 moles reacting = 0.0045moles

the mole ratio in the equation u gave me was 1:1 I2 : S4O6

so therefore there were 0.0045 moles of iodine reacting

therefore using the first equation there were 0.0045 moles of chlorine reacting
since one mole takes up 24dm^3, 0.0045 will take up 24*0.0045 which is 0.108dm^3

I believe.
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john !!
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Oh I read it as S(4)O(6)

Ok, then:

ok.sodium thiosulphate: S4O6 so
45cm^3 = 0.045dm^3
conc. = 0.1M
so there are 0.045*0.1 moles reacting = 0.0045moles

if the ratio of tsulphate to iodine is 2:1 then 2 moles need only 1 iodine to react, or 1 mole of I2 needs 2 of tsulphate to react. so if there are 0.0045 moles of t.sulphate then there will be half the moles of I2 needed

0.0045=2 = 0.00225 moles of iodine

which is 0.00225 moles of chlorine gas reacting, working backwards

which is (going by your scale this time)

25 litres per mole * 0.00225 moles = 0.05625 litres.

lol. did I do any better this time round

edit-im used to using 24dm^3 at rtp rather than 25l
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Sugaray
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Yep! I think so, I found where my mistake was.. but just wondering what the relevance was at the beginning when they said 20 litres of air, contaminated with chlorine gas was bubbled through, is this 20L not used at all? Thanks!!
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Sugaray
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Quick question: when they say 30cm(3) 0.02M KMnO(4) used, the number of moles this is, is it the same as the number of moles of MnO(4)-?
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john !!
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No, because you need to take into account the K+ ions in aqueous form as well, I think. i'm not sure I understand the question.

And in response to the second last one, no you don't use that figure at all, maybe it was a red herring thrown in my the question maker. I can't think what else, because you won't need to know that unless you want the percentage composition by volume of the chlorine in the air mixture.
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Sugaray
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Ok thanks!
Well the second question was:

This question concerns the determination of the amount of preservative, sodium sulphite Na(2)SO(3) in a sample of beefburgers. In an experiment 1kg of meat was boiled with excess dilute HCl (step 1). The SO(2) gas relased was completely absorbed in an excess of dilute aqueous sodium hydroxide (step 2). The resulting solution was then acidified with dilute H(2)SO(4) and titrated with 0.02M KMnO4 soln (step 3): 30cm(3) were required to reach the end point.

step 1: Na(2)SO(3) + 2HCl --> 2NaCl + SO(2) + H(2)O
step 2: SO(2) + 2OH- ---> H(2)O + SO(3)²-
step 3: 5SO(3)²- + 2MnO(4)- + 6H- --> 5SO(4)²- + 2Mn²+ + 3H(2)O

a) How many moles of Na(2)SO(3) are equivalent to 1 mol of MnO(4)-?

I get 2.5?

b) How many moles MnO(4)- were used in the titration?

I get (0.02 x 30)/1000 = 0.0006moles ?

c) moles of Na(2)SO(3) present in 1kg meat?

I'm getting 0.0006/2 x 5 = 0.0015 moles

I'm getting the feeling it's all wrong, but how do you figure the actual moles without the K etc then? Thanks so much for your help ;-)
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Sugaray
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Sorry, again! um in step 1, is it necessary to use an excess of dil HCl to prevent release of Cl gas?? And is it boiled to kill impurities or other substances that can react? Also in step 3, what would be the colour change at the end-point? Is it pink --> colourless? Thanks very much
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Jonatan
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(Original post by Sugaray)
Quick question: when they say 30cm(3) 0.02M KMnO(4) used, the number of moles this is, is it the same as the number of moles of MnO(4)-?
KMnO4 ---> K+ + MnO4-

As you can see from teh above equation. 1 mole of KMnO4 dissloved in water gives exactly 1 mole of MnO4- ions. Btw, KMnO4 is really good at colouring water. Try to put a few barely visible traces of it in a glass and then add some water.
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Jonatan
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(Original post by Sugaray)
Sorry, again! um in step 1, is it necessary to use an excess of dil HCl to prevent release of Cl gas?? And is it boiled to kill impurities or other substances that can react? Also in step 3, what would be the colour change at the end-point? Is it pink --> colourless? Thanks very much
The manganate ion ( MnO(4)- ) gives teh solution a pink/ purple colour (Black if you get much of it , much here being very little). Thsu I think it will be pink --> colourless unless you get some other colour at teh right hand side. (Mn is a transition metal, so it is not unlikely that its ion Mn2+) will colour the solution).
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john !!
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(Original post by Jonatan)
KMnO4 ---> K+ + MnO4-

As you can see from teh above equation. 1 mole of KMnO4 dissloved in water gives exactly 1 mole of MnO4- ions. Btw, KMnO4 is really good at colouring water. Try to put a few barely visible traces of it in a glass and then add some water.
It also goes brown on your finger when you touch it :cool:
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