5.5 g of a mixture of FeSO4⋅7H2O and Fe2(SO4)3⋅9H2O requires 5.4mL of
0.1N KMnO4 solution for complete oxidation. Calculate the number of moles of hydrated Ferric Sulphate in the mixture.
My effort in solving this: I know that I will have to calculate equivalents here, but I'm not sure how to do so using moles. Any hints will do.
N is normality.
Note : I do not want complete solution, rather hints.
[Edited on 19-3-2013 by sankalpmittal]
[Edited on 19-3-2013 by sankalpmittal]Magpie - 19-3-2013 at 11:49
I will take a stab at this based on two assumptions:
1) 0.1N KMnO4 = 0.1M KMnO4
2) The following redox equation applies:
5Fe++ + MnO4- + 8H+ ---> 5Fe+++ + Mn++ + 4H2O
5.4mL of 0.1N KMnO4 = (5.4/1000)(0.1) = 0.0005 mole
This will oxidize (0.0005)(5) moles of Fe++ to produce (0.0005)(5/2) moles of Fe2(SO4)3*9H2O.
MW Fe2(SO4)3*9H20 = 562.02
Therefore g of Fe2(SO4)3*9H2O = (0.0005)(5/2)(562.02)
= 0.703 produced
MW FeSO4*7H2O = 278.02
g of FeSO4*7H2O converted = [2(278.02)/562.02](0.703) =
0.696g
Therefore total wt of Fe2SO4*9H2O now in solution =
5.5g - 0.696g + 0.703 = 5.5g
[Edited on 19-3-2013 by Magpie]woelen - 19-3-2013 at 12:18
The N-notation is used for equivalents of single units.
For redox-systems, a 1 N solution consumes or produces 1 mole of electrons per liter of solution. For acidified MnO4(-), a 1 N solution is a 0.02 M
solution. For acidified Cr2O7(2-), a 1 N solution is a 0.016667 M solution, because 1 ion Cr2O7(2-) consumes 6 electrons in a redox reaction.
For acid-base reactions a 1 N solution consumes or produces 1 mole of H(+) ions. E.g. a 1 N solution of HCl is 1 M, but a 1 N solution of H2SO4 is 0.5
M. A 1 N solution of NaOH is 1 M, a 1 N solution of Ca(OH)2 is 0.5 M.Magpie - 19-3-2013 at 13:33
For redox-systems, a 1 N solution consumes or produces 1 mole of electrons per liter of solution. For acidified MnO4(-), a 1 N solution is a 0.02 M
solution.
Do you mean, instead, a 0.2 M solution?Magpie - 19-3-2013 at 16:04
OK, so I will recalculate this based on the definition of a g-equivalent wt for a redox equation provided by woelen:
Assumptions:
1) 0.1N KMnO4 = 0.02M KMnO4
2) The following redox equation applies:
5Fe++ + MnO4- + 8H+ ---> 5Fe+++ + Mn++ + 4H2O
5.4mL of 0.1N KMnO4 = (5.4/1000)(0.02) = 0.00001 mole
This will oxidize (0.00001)(5) moles of Fe++ to produce (0.00001)(5/2) moles of Fe2(SO4)3*9H2O.
MW Fe2(SO4)3*9H20 = 562.02
Therefore g of Fe2(SO4)3*9H2O = (0.00001)(5/2)(562.02)
= 0.141g produced
MW FeSO4*7H2O = 278.02
g of FeSO4*7H2O converted = [2(278.02)/562.02](0.141) =
0.139g
Therefore total wt of Fe2SO4*9H2O now in solution =
5.5g + 0.141g -139 = 5.5gsankalpmittal - 19-3-2013 at 19:46
Yes, you are correct but question asked for moles and not total mass.
Ok, so all I was having problem was equation and now that you wrote it:
From balanced equation, we have a mole equation :
5*moles of KMnO4 = Moles of FeSO4 hydrated = 5.4*10^-4
Mass of FeSO4 oxidized : 0.15012 g
Mass of Ferric salt in mixture= 5.5-0.15012 = 5.34988 g
Moles of Ferric hydrate already in mixture : 5.34988/562= 0.00952
Ok, the book has the answer 0.0095.. Nearly correct.woelen - 19-3-2013 at 23:43
For redox-systems, a 1 N solution consumes or produces 1 mole of electrons per liter of solution. For acidified MnO4(-), a 1 N solution is a 0.02 M
solution.
Do you mean, instead, a 0.2 M solution?
My mistake! Indeed, I mean 0.2 M instead of 0.02 M.sankalpmittal - 19-3-2013 at 23:58
Nope. 0.02 is correct.
Molarity = Normality/n-factor
Here n-factor is 5 because Mn7+ ------> Mn2+
Thus
Molarity = 0.1/5=0.02 M
Then only I get the answer.Magpie - 20-3-2013 at 06:49
