ATAR Notes: Forum
VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: kriptik on March 28, 2010, 09:06:52 pm
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just wanted to know how to do this for revision
thanks
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hey, i'm quite sure this is how it's done let me know if it's the right answer
EDITED, i realise i made a silly error at the start, here is corrected (with my mistakes crossed out)
first find the total mol of Cr207^2-=0.0100x0.01000=0.000100mol reacted
using the equation to get mol of SO2 reacted x3 = 0.000300mol not to be done yet since we are finding the amount of Cr207 in excess first to find out the amount of it which reacted with it before we find the amount of SO2 .___.
since more solution was used to get the SO2 to react, we use the next part of the question to find out the amount of solution in excess in a titration to then find out the real amount of SO2 reacted.
so n(Fe)=0.0200 x 0.0150 = 0.00300 mol, using equation n(Cr207^2-)=1/6 x N(fe) = 0.0000500mol
now we find out the amount of Cr207^2- reacted with the SO2 which is = amount in total - amount in excess
= 0.000100 - 0.0000500 = 0.00005mol, now since using the original equation n(SO2)=3x n(Cr207 reacted) = 0.00015mol
now since it says escaped "as sulfur dioxide" we must times this by molar mass of SO2, = 0.00015 x 64 = 0.0096g
and for the %= 0.0096x100/.01 = 96%
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I got 48.15%... different from vexx's answer...
And I think you stuffed up somewhere vexx, 0.0200 x 0.0150 doesn't equal 0.00300, but in fact 0.000300 :P
I did whole different steps, but I don't know how to use that special maths text so bear with me.
1. Find initial amount of Cr2O7, so 0.01 x 0.01 = 0.0001mol of initial CrO7
2. Find the unreacted amount of Cr2O7, so 0.015 x .0200= 0.0003mol of Fe that reacted with Cr2O7. Divide by 6 as there are 6 moles of Fe2+ to 1 mole of Cr2O7. 0.0003/6=0.00005mol of unreacted Cr2O7.
3. Find out amount of Cr2O7 reacted with sulfur dioxide- 0.0001-0.00005=0.00005mol
4. Since 0.00005mol of Cr2O7 reacted with 3 moles of SO2, n(SO2)= 0.00005 x 3 = 0.00015mol
5. n(S)= n(SO2)
6. 32.1 x 0.00015 = 0.004815g
7. 0.004815g/0.010g (mass of sulfur) x 100 = 48.15%
8. Sig fig= 48%
Correct me if I'm wrong, I'm a little rusty :P
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Whats the answer by the way? :P
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Whats the answer by the way? :P
sorry i didn't read what i had done, i made a stupid mistake at the start, and it was 96% as it has the amount as sulfur dioxide not as sulfur (so x64 not x32)
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answer should be 48% guys,
thanks for replying :)
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answer should be 48% guys,
thanks for replying :)
ahh that question is confusing then as why does it say "as sulfur dioxide"
oh well:))) no probs hehe.
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answer should be 48% guys,
thanks for replying :)
Woohoo!
No problems :)