z89
- 2
- 0
Actually this is a part of homework. I've found the answer, but I have some problems.
The lab I did in previous lesson is the preparation of K3Fe(C2O4)3.3H2O from Fe(NH4)2(SO4)2.6H2O
In the first step, Iron(II) ammonium sulphate was reacted with oxalic acid dihydrate to form Iron(II) oxalate precipitate.
In the second step, Iron(II) oxalate was oxidized to potassium trioxalatoferrate(III) K3Fe(C2O4)3 by addition of H2O2 and K2C2O4.
In fact, I found link(s) talking about the lab
http://www.chem.umass.edu/genchem/chem112/112_Experiment_1.htm
but I was a bit doubt about the equation mentioned. I observed that there was bubbles formed/gas evolving in step 2. I believed that this is not oxygen, since Fe2+ is oxidized to Fe3+ in this process, H2O2(O: -1 in ON) cannot be oxidized to O2(O: 0) in the same time.
Therefore, it should be CO2, which was oxidized from oxalate, am I right?
And then, why the equation did not mention the oxidization of oxalate... even the release of gas
of course, this problem didn't affect the calculation process. I am just curious ..
Thanks.
The lab I did in previous lesson is the preparation of K3Fe(C2O4)3.3H2O from Fe(NH4)2(SO4)2.6H2O
In the first step, Iron(II) ammonium sulphate was reacted with oxalic acid dihydrate to form Iron(II) oxalate precipitate.
In the second step, Iron(II) oxalate was oxidized to potassium trioxalatoferrate(III) K3Fe(C2O4)3 by addition of H2O2 and K2C2O4.
In fact, I found link(s) talking about the lab
http://www.chem.umass.edu/genchem/chem112/112_Experiment_1.htm
but I was a bit doubt about the equation mentioned. I observed that there was bubbles formed/gas evolving in step 2. I believed that this is not oxygen, since Fe2+ is oxidized to Fe3+ in this process, H2O2(O: -1 in ON) cannot be oxidized to O2(O: 0) in the same time.
Therefore, it should be CO2, which was oxidized from oxalate, am I right?
And then, why the equation did not mention the oxidization of oxalate... even the release of gas
of course, this problem didn't affect the calculation process. I am just curious ..
Thanks.
