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
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
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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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