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

one Fe+2 in Fe(NH4)2(SO4)2*6H20, goes to Fe+3 with 1 electron losttwo oxygens @ -1 each in H2O2, goes to Oxide @ -2 each with 2 electrons takenso balance the give & take # of electrons:2 Fe(NH4)2(SO4)2*6H20 + __H2C2O4 + __K2C2O4 + 1 H2O2 –> __K3Fe(C2O4)3*3H20 + __(NH4)2SO4 + __H2SO4 + H20balance Fe ‘s & NH4 ‘s:2 Fe(NH4)2(SO4)2*6H20 + __H2C2O4 + __K2C2O4 + 1 H2O2 –> 2 K3Fe(C2O4)3*3H20 + 2 (NH4)2SO4 + __H2SO4 + H20balance 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 + H20balance 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 + H20balance 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 H20check 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% H2Obut 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 Kiron = 11.4g @ 55.85 g/mol = 0.204 mol Fewater = 11.0g @ 18g/mol = 0.611 mol H2Ooxalate = 43.1g @ 88 g/mol = 0.490 mol C2O40.882 mol K / 0.204 = 4.32 0.204 mol Fe / 0.204 = 1.000.611 mol H2O / 0.204 = 3.000.490 mol C2O4 / 0.204 = 2.40roughly triple every thing givesK13 Fe 3 (C2O4) 7 . 9 H2Oagain it looks like the % for K % C2O4 might be off=======================... Show More