H2O2 + KMnO4 ---> O2 + MnO2 + KOH + H2O
2007-01-18 00:26:34 · 8 answers · asked by Will 2 in Science & Mathematics ➔ Chemistry
2H2O2 + 2KMnO4 ---> 5/2O2 + 2MnO2 + 2KOH + H2O
OR: 4H2O2 + 4KMnO4 ---> 5O2 + 4MnO2 + 4KOH + 2H2O
2007-01-18 01:37:16 · answer #1 · answered by li mei 3 · 0⤊ 0⤋
you can balance the harder one with oxidation reduction method but here is the easiest one
add prefixes a,b,c,d,e and f to each of the compounds
then for hydrogen you say 2A +2E
like wise you do these for all of them to make i clear for oxygen you do 2a+4b+2c+2d+e+f
then you can make one you the easiest equations one and compete for the others
2007-01-18 04:43:16 · answer #2 · answered by T-bag 3 · 0⤊ 1⤋
4H2O2 + 4KMnO4 -> 4MnO2 + 4KOH + 6H2O + 7O2
2007-01-18 00:48:29 · answer #3 · answered by christopher N 4 · 0⤊ 0⤋
3H202+KMn04----3O2+MnO2+KOH+H20
http://adomas.org/bceq/
http://www.webqc.org/balance.php
http://sciencesoft.at/index.jsp?link=solve&lang=en
2007-01-18 00:32:17 · answer #4 · answered by Natashya K 3 · 0⤊ 1⤋
3H2O2+2KMnO4-----> 3O2+2KMnO2+2KOH+2H2O
2007-01-18 00:40:23 · answer #5 · answered by shirshendu 1 · 0⤊ 0⤋
it is already balanced
2007-01-18 00:31:02 · answer #6 · answered by popo 1 · 0⤊ 2⤋
3H2O2+2KMnO4---------3O2+2MnO2+2KOH+2H2O
2007-01-18 00:45:26 · answer #7 · answered by bashir_khan88 1 · 0⤊ 0⤋
3H2O2+2KMnO4------3O2+2MnO2+2KOH+2H2O
2007-01-18 00:35:31 · answer #8 · answered by amit m 1 · 0⤊ 1⤋