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Topic: How to calculate the number of moles of Fe2+ produced in the reaction?  (Read 28366 times)

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Offline mark1950

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An iron nail was dissolved in excess dilute sulphuric acid to produce 100 cm3 solution. A sample of 10 cm3 of this solution required 4 x 10-4 mol of potassium manganate (VII) for complete reaction.


Fe(s) + H2SO4 (aq)  :rarrow: Fe2SO4 (aq) + H2 (g)
5Fe2+ (aq) + MnO4-  :rarrow: Mn2+ + 4H2O (l) + 5Fe3+ (aq)

If the mass of nail used was 1.45g, calculate the number of moles of Fe2+ produced in the reaction.

The solution they gave was simple:

5 x (4 x 10-4) x 10
= 0.02 mol.

But I don't understand. If 1 mol of MnO4 requires 5 mol of Fe2+, wouldn't it mean that 1 mol of Fe 2+ uses 1/5 mol of MnO4 to produce 1 mol of Fe3+? So, shouldn't it be 4 x 10-4 / 5 instead of 4 x 10-4 x 5?

Also, why do they need to x 10?

My solution was 1.45 / 56 = 0.03 mol since 1 mol of Fe require 1 mol of H2SO4 to produce 1 mol of FeSO4. Why is this not accepted? Thanks.
« Last Edit: October 11, 2009, 09:23:19 AM by mark1950 »

Offline cth

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Re: How to calculate the number of moles of Fe2+ produced in the reaction?
« Reply #1 on: October 11, 2009, 08:51:30 AM »
5Fe2+ (aq) + MnO4-  :rarrow: Mn2+ + 4H2O (l) + 5Fe2+

There is a problem with that equation: the number of electric charges on each side do not agree. +9 on the left, +12 on the right.
And as well, you have Fe2+ on both side. It can't be because it gets oxidised during the reaction.

Offline Borek

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Re: How to calculate the number of moles of Fe2+ produced in the reaction?
« Reply #2 on: October 11, 2009, 08:54:26 AM »
Easy part first:

Quote
Also, why do they need to x 10?

Compare with the question:

Quote
100 cm3 solution. A sample of 10 cm3 of this solution required

But rest is a mess.

5Fe2+ (aq) + MnO4-  :rarrow: Mn2+ + 4H2O (l) + 5Fe2+ (aq)

This is wrong.

Quote
If the mass of nail used was 1.45g, calculate the number of moles of Fe2+ produced in the reaction.

Quote
My solution was 1.45 / 56 = 0.03 mol since 1 mol of Fe require 1 mol of H2SO4 to produce 1 mol of FeSO4. Why is this not accepted? Thanks.

IMHO this is a correct approach, although 0.03 is not the correct (numerical) answer. You should use three significant figures.

Quote
The solution they gave was simple:

5 x (4 x 10-4) x 10
= 0.02 mol.

But I don't understand. If 1 mol of MnO4 requires 5 mol of Fe2+, wouldn't it mean that 1 mol of Fe 2+ uses 1/5 mol of MnO4 to produce 1 mol of Fe3+? So, shouldn't it be 4 x 10-4 / 5 instead of 4 x 10-4 x 5?

No, you have it reversed. 1 mole of permanganate oxidizes 5 moles of iron, so to calculate moles of iron you have to multiply moles of permanganate by 5.

But I have no idea why they won't you to go through permanganate, if they have given mass of the nail - unless it was not iron, but some alloy. Have you quoted question exactly as it was worded?
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Offline mark1950

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Re: How to calculate the number of moles of Fe2+ produced in the reaction?
« Reply #3 on: October 11, 2009, 09:37:05 AM »
Sorry about the equation. it was supposed to be

5Fe2+ (aq) + MnO4-  :rarrow: Mn2+ + 4H2O (l) + 5Fe3+ (aq) Typo error.

Quote
No, you have it reversed. 1 mole of permanganate oxidizes 5 moles of iron, so to calculate moles of iron you have to multiply moles of permanganate by 5.

I'm having a hard time trying to grasp the logic here. I've always thought that 1 kg of rice can be moulded into 5 packets of rice and each will contain 0.2 kg of rice. Hence, I'm applying this to this chemistry thing. Since 1 mol of MnO4 requires 5 mole of Fe2+, 1 mol of Fe2+ would need 1/5 mole of MnO4.

Quote
100 cm3 solution. A sample of 10 cm3 of this solution required

This would mean that the number of moles of iron in 100 cm3 of FeSO4 is 10 times the number of mole in 10cm3 of FeSO4. So, you're saying that the mole varies with the volume of solution? How does it happens? I know that concentration varies with the volume of solution.

Offline Borek

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Re: How to calculate the number of moles of Fe2+ produced in the reaction?
« Reply #4 on: October 11, 2009, 09:49:28 AM »
I'm having a hard time trying to grasp the logic here. I've always thought that 1 kg of rice can be moulded into 5 packets of rice and each will contain 0.2 kg of rice. Hence, I'm applying this to this chemistry thing. Since 1 mol of MnO4 requires 5 mole of Fe2+, 1 mol of Fe2+ would need 1/5 mole of MnO4.

Correct. 1 mole of Fe2+ requires 1/5 mole of permanganate, or 1/5 mole permanganate oxidises 1 mole of irom. 5x1/5=1. You have to multiply moles of permanganate by 5.

Quote
This would mean that the number of moles of iron in 100 cm3 of FeSO4 is 10 times the number of mole in 10cm3 of FeSO4. So, you're saying that the mole varies with the volume of solution? How does it happens? I know that concentration varies with the volume of solution.

Not mol varies but number of moles varies. If you have 1 kg of rice and you take 1/10th of that you don't expect to have 1 kg of rice, but 0.1 kg of rice, don't you?
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Offline mark1950

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Re: How to calculate the number of moles of Fe2+ produced in the reaction?
« Reply #5 on: October 11, 2009, 10:15:51 AM »
Uhuh...so number of moles varies...

Quote
Correct. 1 mole of Fe2+ requires 1/5 mole of permanganate, or 1/5 mole permanganate oxidises 1 mole of irom. 5x1/5=1. You have to multiply moles of permanganate by 5.
Oh! I see your point. It says,

Quote
A sample of 10 cm3 of this solution required 4 x 10-4 mol of potassium manganate (VII) for complete reaction.

So that means, 1 mole of Fe2+ in FeSO4 require 4 x 10-4 mol of potassium manganate to be oxidised to 1 mole of Fe3+. Since the ionic equation shows 5 moles, than we multiply by 5...and we get the number of moles of Fe2+ in 10cm3 of solution. Then we x by 10 to get the number of moles in 100cm3 of solution, right?

Offline Borek

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Re: How to calculate the number of moles of Fe2+ produced in the reaction?
« Reply #6 on: October 11, 2009, 10:22:13 AM »
So that means, 1 mole of Fe2+ in FeSO4 require 4 x 10-4 mol of potassium manganate

No, it means that SOME amount (as of yet unknown) requires 4x10-4 moles of permanganate.

Otherwise you seem to be on the right track.
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