Redox Titration question
"In the presence of dilute sulphuric acid, manganate( VII) ions oxidise ethanedioate ions to carbon dioxide.
A 1.35 g sample of hydrated iron(II) ethanedioate, FeC2O4.xH2O, was dissolved in dilute sulphuric acid and made up to 250 cm3 in a volumetric flask. A 25.0 cm3 sample of the solution was titrated at 70°C with 0.0200 M potassium manganate( VII) solution. The titration required a volume of 22.5 cm3 to produce the first permanent pink colour. Calculate the value of x."
I'm struggling to calculate an answer to this question with the Mr of hydrated iron (II) ethanedioate that I find being smaller than the Mr of dehydrated Iron (II) Ethanedioate.
A 1.35 g sample of hydrated iron(II) ethanedioate, FeC2O4.xH2O, was dissolved in dilute sulphuric acid and made up to 250 cm3 in a volumetric flask. A 25.0 cm3 sample of the solution was titrated at 70°C with 0.0200 M potassium manganate( VII) solution. The titration required a volume of 22.5 cm3 to produce the first permanent pink colour. Calculate the value of x."
I'm struggling to calculate an answer to this question with the Mr of hydrated iron (II) ethanedioate that I find being smaller than the Mr of dehydrated Iron (II) Ethanedioate.
first you divide 22.5 by 1000
multiply the answer of the above by 0.02
multiply the answer for the last step by 10
after that, do 1.35 divided by the answer of the last step
the answer should be the MR of the hydrated version
next take away the mr of Fec2o4 from the MR from the last step
now you are left with the MR of H2O molecules
divide the MR by 18 , which is the MR of h2O.
The answer is the value of X
multiply the answer of the above by 0.02
multiply the answer for the last step by 10
after that, do 1.35 divided by the answer of the last step
the answer should be the MR of the hydrated version
next take away the mr of Fec2o4 from the MR from the last step
now you are left with the MR of H2O molecules
divide the MR by 18 , which is the MR of h2O.
The answer is the value of X
Original post by Joe Walton
first you divide 22.5 by 1000
multiply the answer of the above by 0.02
multiply the answer for the last step by 10
after that, do 1.35 divided by the answer of the last step
the answer should be the MR of the hydrated version
next take away the mr of Fec2o4 from the MR from the last step
now you are left with the MR of H2O molecules
divide the MR by 18 , which is the MR of h2O.
The answer is the value of X
multiply the answer of the above by 0.02
multiply the answer for the last step by 10
after that, do 1.35 divided by the answer of the last step
the answer should be the MR of the hydrated version
next take away the mr of Fec2o4 from the MR from the last step
now you are left with the MR of H2O molecules
divide the MR by 18 , which is the MR of h2O.
The answer is the value of X
Where do you get 10 from?
C2O42- -> 2CO2 + 2e-
MnO4- + 8H+ 5e- -> Mn2+ + 4H2O
5C2O4 + 2MnO4- + 16H+ -> 2Mn2+ + 8H2O + 10CO2
Doesn't this half equation make the mole ratio between MnO4- and C2O4 2:5 meaning you'd multiply your answer by 2.5 instead of 10?
Original post by crazychameleon
Where do you get 10 from?
C2O42- -> 2CO2 + 2e-
MnO4- + 8H+ 5e- -> Mn2+ + 4H2O
5C2O4 + 2MnO4- + 16H+ -> 2Mn2+ + 8H2O + 10CO2
Doesn't this half equation make the mole ratio between MnO4- and C2O4 2:5 meaning you'd multiply your answer by 2.5 instead of 10?
C2O42- -> 2CO2 + 2e-
MnO4- + 8H+ 5e- -> Mn2+ + 4H2O
5C2O4 + 2MnO4- + 16H+ -> 2Mn2+ + 8H2O + 10CO2
Doesn't this half equation make the mole ratio between MnO4- and C2O4 2:5 meaning you'd multiply your answer by 2.5 instead of 10?
MnO4- + 5Fe2+ + 8H+ -> Mn2+ + 4H2O + 5Fe3+
5FeC2O4 + 3MnO4- + 24H+ -> 5Fe3+ + 3Mn2+ + 12H2O + 10CO2
Find number of moles of MnO4-
Find number of moles of FeC2O4 in 25cm3
Find number of moles of FeC2O4 in 250cm3 (multiply above answer by 10)
Find molar mass of FeC2O4.xH2O
Find value of x
There exists alternative methods.
The value of x is 2.01
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