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A

Anonymous

Jan 02, 2025

Find empirical for KxFey(C2O4)z·nH2O and balance equation?

I have the mass percentages of the following ions:

  • Potassium: 34.5%
  • Iron: 11.4%
  • Water: 11.0%
  • Oxalate: 43.1%

How do I determine the empirical formula for KxFey(C2O4)z·nH2O based on these percentages? Additionally, I would greatly appreciate assistance in balancing the following chemical equation:

Fe(NH4)2(SO4)2·6H2O + H2C2O4 + K2C2O4 + H2O2 → K3Fe(C2O4)3·3H2O + (NH4)2SO4 + H2SO4 + H2O.

1 Answers

A
Anonymous

Feb 11, 2025

one Fe+2 in Fe(NH4)2(SO4)2*6H20, goes to Fe+3 with 1 electron lost

two oxygens @ -1 each in H2O2, goes to Oxide @ -2 each with 2 electrons taken

so balance the give & take # of electrons:

2 Fe(NH4)2(SO4)2*6H20 + __H2C2O4 + __K2C2O4 + 1 H2O2 –> __K3Fe(C2O4)3*3H20 + __(NH4)2SO4 + __H2SO4 + H20

balance Fe ‘s & NH4 ‘s:

2 Fe(NH4)2(SO4)2*6H20 + __H2C2O4 + __K2C2O4 + 1 H2O2 –> 2 K3Fe(C2O4)3*3H20 + 2 (NH4)2SO4 + __H2SO4 + H20

balance K’s & SO4 ‘s:

2 Fe(NH4)2(SO4)2*6H20 + __H2C2O4 + 3 K2C2O4 + 1 H2O2 –> 2 K3Fe(C2O4)3*3H20 + 2 (NH4)2SO4 + 2 H2SO4 + H20

balance C2O4 ‘s:

2 Fe(NH4)2(SO4)2*6H20 + 3 H2C2O4 + 3 K2C2O4 + 1 H2O2 –> 2 K3Fe(C2O4)3*3H20 + 2 (NH4)2SO4 + 2 H2SO4 + H20

balance H2O2 redoxing O’s into H2O:

2 Fe(NH4)2(SO4)2*6H20 + 3 H2C2O4 + 3 K2C2O4 + 1 H2O2 –> 2 K3Fe(C2O4)3*3H20 + 2 (NH4)2SO4 + 2 H2SO4 + 2 H20

check on your work, is that H’s should already be balanced:

3 H2C2O4 + 1 H2O2 –> 2 H2SO4 + 2 H20 (8 H’s –>8 H’s)

it’s done @

2 Fe(NH4)2(SO4)2*6H20 + 3 H2C2O4 + 3 K2C2O4 + 1 H2O2 –> 2 K3Fe(C2O4)3*3H20 + 2 (NH4)2SO4 + 2 H2SO4 + 2 H20

=========================================

Find empirical for KxFey(C2O4)z·nH2O

something seems off, because if you wanted K3Fe(C2O4)3*3H20

it is 11.4 % Fe

& it is 11.0% H2O

but it is 23.9% K

& 53.8 % C2O4

————————–

so let’s seee want we can get:

potassium = 34.5g @ 39.1 g/mol K = 0.882 mol K

iron = 11.4g @ 55.85 g/mol = 0.204 mol Fe

water = 11.0g @ 18g/mol = 0.611 mol H2O

oxalate = 43.1g @ 88 g/mol = 0.490 mol C2O4

0.882 mol K / 0.204 = 4.32

0.204 mol Fe / 0.204 = 1.00

0.611 mol H2O / 0.204 = 3.00

0.490 mol C2O4 / 0.204 = 2.40

roughly triple every thing gives

K13 Fe 3 (C2O4) 7 . 9 H2O

again it looks like the % for K % C2O4 might be off

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