CHEMISTRY HELP!!!!!
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Help please! I've tried so hard to come out with the balanced equations but to no avail );
A metallic salt containing the anion IxCly- decomposed when heated to form iodine vapour and the metallic chloride. When 0.270 g of this compund was heated, the iodine evolved was bubbled into excess potassium iodide solution.
The dissolved iodine required 21.30 cm3 of 0.0600 mol dm-3 aqueous sodium thiosulfate for complete reaction. When the residue was dissolved in water, the resulting solution required 6.40 cm3 of 0.100 mol dm-3 aqueous silver nitrate for complete reaction.
Determine the value of x and y in IxCly-.
Hence, determine the identity of the metallic element.A new question:1.5 g of an iron wire was dissolved in excess dilute sulphuric acid and the solution was made up to 200 cm3. 25.0 cm3 of this solution needed 25.45 cm3 of a 0.0188 mol dm-3 potassium manganate (VII)solution for oxidation. Calculate the percentage of iron in the iron wire.What I got (please correct me if I am wrong):
Fe2+ => Fe3+ + e-
MnO4- + 8H+ + 5e- => Mn2+ + 4H2OMnO4- + 8H+ + 5Fe2+ => Mn2+ + 4H2O + 5Fe3+No of moles of MnO4- = 25.45/1000 * 0.0188 = 0.00047846 mol
Mole ratio, MnO4- : Fe2+ = 1 : 5
No of moles of Fe2+ in 25.0 cm3 = 0.00047846 * 5 = 0.0023923 mol
No of moles of Fe2+ in 200 cm3 = 0.0023923/25.0 * 200 = 0.019138 mol
Mass of Fe2+ = 0.019138 ( 55.8 + 32.1 + 16.0 *4) = 2.9071 gI'm stuck here already. Cos apparently my answer for Fe2+ is more than what is provided.Your help is very much appreciated. Thank you
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I2 + 2Na2S2O3 ---> 2NaI + Na2S4O6
Amt of I2 needed = 21.3/1000 x 0.06 x 0.5 = 6.39 x 10^-4 mol
Amt of I = 2 x 6.39 x 10^-4 = 1.278 x 10^-3 mol
Amt of Cl- needed = 6.4/1000 x 0.1 = 6.4 x 10^-4 mol
Therefore ratio of I : Cl = 2:1
then ion is I2Cl-
0.270 g of compound = 0.00064 mol
Molar mass of compound = 421.88 g per mol
molar mass of M = 421.88 - 254 - 35.5 = 132.4
Closest match is Cs (132.9)
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OMGGGGG, thanksthanksthanksthanksthanks!
You've saved my life.
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I've only one more comment to add, that the anionic species I2Cl- is unusual and rare. If it does indeed exist (can anyone reference to it?), the Kekule structure is probably simlar to that of the tri-iodide uninegative anion, I3- (ie. the central halogen atom with the negative formal charge has 3 lone pairs and 2 bond pairs; the terminal halogen atoms have 3 lone pairs and 1 bond pair, each).
Regarding what 'Jodar' wrote erroneously on another forum (on which dyingsoul92 also posted this same question) :
>>> So the empirical formula is I2Cl-. Since Cl is more electronegative than I, I will be positive in this case, while Cl is -1. So the total charge of the ion has to be -1. Which fulfills that stated in the question above: Anion IxCly -1. However, if the formula of the ion is to be I2Cl4, then the charge will have to be -2. Which does not fulfill the IxCly -1 stated in the question. Thus, the ONLY formula of the anion which satisfy it is I2Cl -1. So x = 2 and y = 1. <<<
The ionic charge of a polyatomic ion is the sum of all formal charges within. Electronegativity does not affect the formal charges (and hence overall ionic charge). Electronegativity only affects the oxidation state (O.S.) of the atoms within the ion.It's ok if some of you reading this don't fully understand what I just said. You will, eventually. (hopefully!!!)
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I'm so sorry but I've got a question to ask again.
Now that I know the anion is I2Cl- and assuming the metal is A with a charge of a.
The very first equation i get is:
A(I2Cl)a => aI2 + ACla
The number of moles of A(I2Cl)a would be 0.000639 mol.
Assuming the Ar of A is y,
number of moles of A(I2Cl)a = Mass of A(I2Cl)a / Molecular Mass of A(I2Cl)a
0.000639 = 0.270 / ( y + 2a x 127 + 35.5a)
The question is i don't know a, so how do I go on?With reference to Chemfreak022, how do you know that you just only 1?Pardon me for my stupidity. -
It was already determined that there is 1 chlorine per molecule of the compound so the metal has to be a A+ to get ACl so you should not need the a for the charge.
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A new question:
1.5 g of an iron wire was dissolved in excess dilute sulphuric acid and the solution was made up to 200 cm3. 25.0 cm3 of this solution needed 25.45 cm3 of a 0.0188 mol dm-3 potassium manganate (VII)solution for oxidation. Calculate the percentage of iron in the iron wire.
What I got (please correct me if I am wrong):
Fe2+ => Fe3+ + e-
MnO4- + 8H+ + 5e- => Mn2+ + 4H2OMnO4- + 8H+ + 5Fe2+ => Mn2+ + 4H2O + 5Fe3+
No of moles of MnO4- = 25.45/1000 * 0.0188 = 0.00047846 mol
Mole ratio, MnO4- : Fe2+ = 1 : 5
No of moles of Fe2+ in 25.0 cm3 = 0.00047846 * 5 = 0.0023923 mol
No of moles of Fe2+ in 200 cm3 = 0.0023923/25.0 * 200 = 0.019138 mol
Mass of Fe2+ = 0.019138 ( 55.8 + 32.1 + 16.0 *4) = 2.9071 gI'm stuck here already. Cos apparently my answer for Fe2+ is more than what is provided.
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Originally posted by dyingsoul92:
A new question:
1.5 g of an iron wire was dissolved in excess dilute sulphuric acid and the solution was made up to 200 cm3. 25.0 cm3 of this solution needed 25.45 cm3 of a 0.0188 mol dm-3 potassium manganate (VII)solution for oxidation. Calculate the percentage of iron in the iron wire.
What I got (please correct me if I am wrong):
Fe2+ => Fe3+ + e-
MnO4- + 8H+ + 5e- => Mn2+ + 4H2OMnO4- + 8H+ + 5Fe2+ => Mn2+ + 4H2O + 5Fe3+
No of moles of MnO4- = 25.45/1000 * 0.0188 = 0.00047846 mol
Mole ratio, MnO4- : Fe2+ = 1 : 5
No of moles of Fe2+ in 25.0 cm3 = 0.00047846 * 5 = 0.0023923 mol
No of moles of Fe2+ in 200 cm3 = 0.0023923/25.0 * 200 = 0.019138 mol
Mass of Fe2+ = 0.019138 ( 55.8 + 32.1 + 16.0 *4) = 2.9071 gI'm stuck here already. Cos apparently my answer for Fe2+ is more than what is provided.
Correct until the last step.
Sample mass of Fe = no. of moles x molar mass = 0.019138 x 55.8 = 1.0679g
Percentage of iron = mass of iron / mass of wire = 1.0679 / 1.5 = 0.7119 = 71.19% = 71.2% (3 sig fig)
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That's what i thought so too.
But the final answer is supposed to be 89.0% -
Originally posted by dyingsoul92:That's what i thought so too.
But the final answer is supposed to be 89.0%
Assuming your no. of moles of Fe2+ in 200cm3 are correct (I did not check your previous calculations for errors), then the last step I showed you is correct. Don't blindly trust answers given. What's important is you know what's going on (ie. that you know your concepts) and you know how to deal with questions like these. From your steps above, you're all right (except for the last step you did, calculating sample mass of iron(II) sulfate which is irrelevant). -
1. NH2OH is oxidised by Fe3+ which itself is reduced to Fe2+.
how to write the balanced equation..
i know how to do the Fe3+ to Fe2+ but don't know how to do the NH2OH part.
2. The amt of chloric(I) acid can be determined by adding postassium iodide to a sample of water and then titrating the aq iodine produced by the oxidation of the iodide ion by chloric(I) acid with sodium thiosulphate solution.
HOCl + (H+) + (2I-) + -> I2 + (Cl-) + H20
50 cm3 of water was treated with an excess of potassium iodide and then titrated with thiosulphate in the presense of iodine indicator. 15.80 cm3 of 0.0010 moldm-3 of sodium thiosulphate solution were needed for complete reaction. Calculate the concentration of the chloric(I) acid in the water sample.
this question i'm don't understand it. for the 2nd para of the question. How to write out the equation?
thanks(:
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Q2. 2 S2O32−(aq) + I2(aq) → S4O62−(aq) + 2 I−(aq)
(Note, usually this is a common eqn (well at least for me) of thiosulphate going along with the 2I- => I2 + 2e)
i Guess u know how to do the next part? just find the amt in the redox eqn and sub back to the first eqn involving HOCl
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Originally posted by blueftw:
1. NH2OH is oxidised by Fe3+ which itself is reduced to Fe2+.
how to write the balanced equation..
i know how to do the Fe3+ to Fe2+ but don't know how to do the NH2OH part.
2. The amt of chloric(I) acid can be determined by adding postassium iodide to a sample of water and then titrating the aq iodine produced by the oxidation of the iodide ion by chloric(I) acid with sodium thiosulphate solution.
HOCl + (H+) + (2I-) + -> I2 + (Cl-) + H20
50 cm3 of water was treated with an excess of potassium iodide and then titrated with thiosulphate in the presense of iodine indicator. 15.80 cm3 of 0.0010 moldm-3 of sodium thiosulphate solution were needed for complete reaction. Calculate the concentration of the chloric(I) acid in the water sample.
this question i'm don't understand it. for the 2nd para of the question. How to write out the equation?
thanks(:
On Qn#1
Hydroxyamine is oxidized to nitrite ion aka nitrate(III) ion.
On Qn#2
Edited/added : Eyelessz has already posted (correctly) to blueftw, in the time I was typing out this post.
>>> 50 cm3 of water CONTAINING AN UNKNOWN AMOUNT OF CHLORIC(I) ACID was treated with an excess of potassium iodide and then titrated with thiosulphate in the presense of iodine indicator. 15.80 cm3 of 0.0010 moldm-3 of sodium thiosulphate solution were needed for complete reaction. Calculate the concentration of the chloric(I) acid in the water sample. <<<
There, added the words in bold for you, in case the original question's phrasing confused you.
The (stoichiometric) chain going through the two redox equations is as follows :
However much chloric(I) acid aka hypochlorous acid you have, will result in however much iodide ions oxidized to iodine molecules. (Redox equation provided by question).
However much iodine molecules you get as a result of the above, is found by however much thiosulfate ions you required for complete reduction of any iodine present (starch is usually used as an indicator).
You should be familiar (at 'A' levels and even at 'O' levels) that thiosulfate ions (S2O3 2-) are oxidized to tetrathionate ions (S4O6 2-), when they carry out their reducing work (in this context, reducing iodine to iodide).
Are you able to follow the (stoichiometric) chain through the two redox reactions?
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i can do the questions already (: thanks.
anyone can show a website to self-learn how to write out the spdf notation for the Electronic Configuration?topic is atomic structure,
thanks
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spdf isn't hard.
s just means your group 1 and 2 elements, p your group 3-8, d your transition and those at the bottom which i don't think jc will ever test.
Only exception is the 4th and 9th element in the d block which gives s1d5/s1d10 instead of s2d4/s2d9. All the rest just follows the order.
Egs
Na -> 1s2 2s2 2p6 3s1
Cl -> 1s2 2s2 2p6 3s2 3p5
Cu -> 1s2 2s2 2p6 3s2 3p6 3d10 4s1 (NOT 3d9 4s2)U can also skip writing the nobel gas part of the configuration by writing
Na -> [Ne] 3s1
Cl -> [Ne] 3s2 3p5
Cu -> [Ar] 3d10 4s1 (NOT 3d9 4s2)