help with how to do this question that came up in edexcel unit 5 Watch

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Report Thread starter 11 years ago
help with this q

A 1.00 g sample of a metal alloy that contains chromium was converted into
250 cm3 of an acidified solution of potassium dichromate(VI).
25.0cm3 of this solution was added to an excess of potassium iodide solution.
2– + 6I– + 14H+ → 2Cr3+ + 3I2 + 7H2O
The iodine liberated was titrated with 0.100 mol dm–3 sodium thiosulphate
+ 2S2O3
2– → 2I– + S4O6
The mean (average) titre was 37.2 cm3.
Calculate the amount (moles) of iodine liberated and hence the percentage, by
mass, of chromium in the alloy.
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Report 11 years ago
Wow, this was a longer answer than I was expecting to give :p: I think this is right... I'm sure someone will correct me if not. ONWARDS!

You kind of have to work backwards, so first of all work out how many moles of thiosulphate ions were in the titre.

Using Concentration = number of moles / volume,

0.100 mol dm-3 = (number of moles) / (37.2/1000) dm-3 (divide by 1000 to change from cm-3 to dm-3)

So there were 0.100 x (37.2/1000) = 0.00372 moles of thiosulphate in the titre.

From the second balanced equation, every 1 mole of iodine reacts with 2 moles of thiosulphate ions. So, however many moles of thiosulphate were in the titre reacted with half that number of moles of iodine.
So, the iodine liberated was 0.00372 / 2 = 1.86 x 10^-3 moles.

Right, now, from the first balanced equation, 1 mole of Cr2O7 ions liberated 3 moles of I2. That means that, however many moles of iodine were liberated, you started with one third that number of moles of Cr2O7 ions.
So, having just calculated that 1.86 x 10^-3 moles of iodine were liberated, you divide that by three to find that there must have been 6.2 x 10^-4 moles Cr2O7 ions initially.

In every one mole of Cr2O7 ions, there are TWO moles of chromium ions, so in 6.2 x 10^-4 moles of Cr2O7 ions, there are (6.2 x 10^-4) x 2 = 1.24 x 10^-3 moles chromium ions, so that's the number of moles of chromium you started with.

BUT the question says that you only used 25 cm^3 of the 250 cm^3 that you made up. So, in the original alloy, there was actually 10 times this number of moles, so we started with 1.24 x 10^-2 moles of chromium.

1 mole of chromium weighs 52 g (its atomic mass), so, considering we've only got 1.24 x 10^-2 moles, that's only going to weigh 52 x (1.24 x 10^-2) = 0.6448 g chromium. (Mass = number of moles x molar mass.)

The whole alloy weighed 1g, and 0.6448g out of this was chromium, so that's (0.6448/1)x100 = 64.48 % = 64.5 % (to 3 s.f.)

Phew! I really hope that's right now...

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