Redox titrations help!!!!

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doglover123
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#1
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I was doing alright until I came across this question . i worked out the moles of (NH4)Cr2O7 as being 4x10-4, but I've got no idea where to go from there. Thanks in advance
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h3rmit
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(Original post by doglover123)
I was doing alright until I came across this question . i worked out the moles of (NH4)Cr2O7 as being 4x10-4, but I've got no idea where to go from there. Thanks in advance
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Have you done the equations?
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doglover123
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(Original post by h3rmit)
Have you done the equations?
Cr2O7 + 6Fe + 14H = 2Cr + 6Fe + 7H2O

then i used the mole ratios to calculate the moles of ammonium dichromate as being 4x10-4. Is there another equation I need?
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h3rmit
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(Original post by doglover123)
Cr2O7 + 6Fe + 14H = 2Cr + 6Fe + 7H2O

then i used the mole ratios to calculate the moles of ammonium dichromate as being 4x10-4. Is there another equation I need?
I meant the ionic half equations, but the overall equation is better.

Do you have any other ideas at all?

You also need the moles of iron sulphate.
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h3rmit
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(Original post by doglover123)
Cr2O7 + 6Fe + 14H = 2Cr + 6Fe + 7H2O

then i used the mole ratios to calculate the moles of ammonium dichromate as being 4x10-4. Is there another equation I need?
Also, just to check, did you calculate the moles of dichromate via the 0.233g?
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doglover123
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(Original post by h3rmit)
Also, just to check, did you calculate the moles of dichromate via the 0.233g?
No - I worked out the moles of dichromate that reacted wwith the acidic iron sulphate solution, because we were given the conc and volume of that.

I know the moles of iron sulphate that reacted was 2.4x10-3 - but ive got no clue where to go from there
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h3rmit
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(Original post by doglover123)
No - I worked out the moles of dichromate that reacted wwith the acidic iron sulphate solution, because we were given the conc and volume of that.
Good, now you know how many actual moles of dichromate there are. You then calculate how many moles of dichromate there would be if the 0.233g was pure. Then, the difference in moles tells you your number of moles of ammonium chloride.
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jordenevans12
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Is it possible to use moles = mass / molar mass with 4x10-4 and 0.223g
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doglover123
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(Original post by h3rmit)
Good, now you know how many actual moles of dichromate there are. You then calculate how many moles of dichromate there would be if the 0.233g was pure. Then, the difference in moles tells you your number of moles of ammonium chloride.
U BEAUTY THANK YOU!!!! I got the answer!
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doglover123
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(Original post by jordenevans12)
Is it possible to use moles = mass / molar mass with 4x10-4 and 0.223g
Yeah you can, then subtract that value from the original mass (0.223g) to work out how many moles of ammonium chloride reacts & you can work out the ratio from there
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h3rmit
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(Original post by jordenevans12)
Is it possible to use moles = mass / molar mass with 4x10-4 and 0.223g
Since you have a number of moles and a mass, you could only use that to calculate the molar mass. And, since the 4x10^-4 is only part of the 0.233g, you'd be calculating the molar mass of a tiny part of the 0.233g.
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jordenevans12
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(Original post by h3rmit)
Since you have a number of moles and a mass, you could only use that to calculate the molar mass. And, since the 4x10^-4 is only part of the 0.233g, you'd be calculating the molar mass of a tiny part of the 0.233g.
Yeah, we have the molar mass already from the formula using the periodic table. Find the mass of 4x10^-4 moles of ammonium chromate and subtract from 0.233g then we will have a ratio
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