K4Fe(CN)6 + Ce(NO3)4 + KOH = Ce(OH)3 + Fe(OH)3 + H2O + K2CO3 + KNO3 please balance?

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K4Fe(CN)6 + Ce(NO3)4 + KOH = Ce(OH)3 + Fe(OH)3 + H2O + K2CO3 + KNO3 please balance?

use method other than hit and trail. please explain in detail.
how to balance an equation in which an element makes two or more compounds.
do you have any idea about 'n' factor balancing. please help

2007-05-16 21:43:56 · 4 answers · asked by mathiphy 2 in Science & Mathematics ➔ Chemistry

4 answers

K4[Fe(CN)6] + Ce(NO3)4 + 18 KOH = Ce(OH)3 + Fe(OH)3 + 9 H2O + 6 K2CO3 + 10 KNO3

2007-05-17 06:02:34 · answer #1 · answered by ag_iitkgp 7 · 0⤊ 0⤋

K4Fe(CN)6 + Ce(NO3)4 + KOH = Ce(OH)3 + Fe(OH)3 + H2O + K2CO3 + KNO3

2014-05-05 20:04:06 · answer #2 · answered by Anonymous · 0⤊ 0⤋

K2CO3+H2O+Fe(OH)3+Ce(NO)4+KOH→K4Fe(CN)6+Ce(OH)3+KNO3

2016-08-29 08:25:29 · answer #3 · answered by azita 1 · 0⤊ 0⤋

258KOH + K4Fe(CN)6 + 61Ce(NO3)4 = 61Ce(OH)3 + Fe(OH)3 + 36H2O + 6K2CO3 + 250KNO3

2014-05-11 18:46:51 · answer #4 · answered by Arijit 1 · 0⤊ 0⤋