balancing elemental equation.... (extremely hard)?

Question

K4Fe(CN)6 + KMnO4 + H2SO4 -> KHSO4 + Fe2(SO4)3 + MnSO4 + HNO3 + CO2 + H2O ......... al numbers are the small numbers after the element not the big number like 3SO but like H2O....... 1 before this was incorrect

Answer 1

If you want, I can elaborate with the exact methodology, but basically how you do a problem like this is to just keep isolating groups that have to be the same on both sides.

For example, you can observe that you only have one source of Fe on both sides, so you MUST have twice as much of the reactant molecule as the product one. You can also note that none of the sulfates ever break up, so that leaves you with only one source of sulfate and one source of oxygen among the reactants. That helps too! You likewise have only one source of carbon and nitrogen and only one product each for those. You can joggle these all around and get the answer:

(Updated from previous, slightly incorrect answer)

10 K4Fe(CN)6 + 122 KMnO4 + 299 H2SO4 -> 162 KHSO4 + 5 Fe2(SO4)3 + 122 MnSO4 + 60 HNO3 + 60 CO2 + 188 H2O

I did a couple checks and everything REALLY seems to work out this time. If you don't have to have the answer (like me), doing one can be fun like doing a puzzle! Thanks! ( :

Answer 2

K4Fe(CN)6 + KMnO4 + H2SO4 -> KHSO4 + Fe2(SO4)3 + MnSO4 + HNO3 + CO2 + H2O Mn is going from +7 to +2. help. C is going from +2 to +4. Oxidation. N is going from -3 to +5. Oxidation. Fe is going from +2 to +3. Oxidation. for each K4Fe(CN)6 that's oxidized, that's +61. for each Mn that's decreased, that's -5. subsequently, the Mn to CN ratio is 61 to 5. 10 K4Fe(CN)6 + 122 KMnO4 + 299 H2SO4 -> 162 KHSO4 + 5 Fe2(SO4)3 + 122 MnSO4 + 60 HNO3 + 60 CO2 + 188 H2O examine: that's 162 ok on both area, 10 Fe, 60 C, 60 N, 122 Mn, 598 H, 299 S, and a whopping 1684 O.

Answer 3

This is a redox reaction. You need to identify all the atoms that change oxidation number and write all the half reactions. This question is particularly difficult only because there are more than two atoms changing oxidation numbers.

Once you write the half equations use the normal method of balancing redox equations.