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
2006-09-30 10:12:03 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
Not that hard. This is an oxidation-reduction reaction.
10 K4Fe(CN)6 + 122 KMnO4 + 299 H2SO4 -->
162 KHSO4 + 5 Fe2(SO4)3 + 122 MnSO4 + 60 HNO3 + 60 CO2 + 188 H2O
2006-09-30 10:15:18 · answer #1 · answered by Richard 7 · 69⤊ 0⤋
K4Fe(CN)6 + KMnO4 + H2SO4 -> KHSO4 + Fe2(SO4)3 + MnSO4 + HNO3 + CO2 + H2O
Mn goes from +7 to +2. Reduction.
C goes from +2 to +4. Oxidation.
N goes from -3 to +5. Oxidation.
Fe goes from +2 to +3. Oxidation.
for every K4Fe(CN)6 that's oxidized, that's +61.
for every Mn that's reduced, that's -5.
Thus, 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
Check: That's 162 K on each side, 10 Fe, 60 C, 60 N, 122 Mn, 598 H, 299 S, and a whopping 1684 O.
2006-09-30 13:15:10 · answer #2 · answered by Mr. E 5 · 1⤊ 0⤋
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2016-11-25 19:06:08 · answer #3 · answered by hyre 4 · 0⤊ 0⤋
Drive a nail in a board with point up try to get the equation to balance on the point. When it does not fall in either direction you have balanced the equation.
2006-09-30 10:50:32 · answer #4 · answered by jess g 3 · 0⤊ 0⤋
im not sure if theres a mistake but CN is not present on the right handside of the equation
2006-09-30 10:20:27 · answer #5 · answered by Anonymous · 0⤊ 0⤋