Предмет: Химия, автор: bomber8725

Закончите уравнения реакций, расставив коэффициенты методом электронно-ионного (или электронного) баланса: MnCl2 + KClO + KOH =K2MnO4 + KCl + H2O

Ответы

Автор ответа: Аноним

MnCl2 + 2KClO + 4KOH = K2MnO4 + 4KCl + 2H2O:

Mn2+ -4e- + 8OH- = (MnO4)2- + 4H2O | 4 | 4 | 1 |

(ClO)- +2e- + H2O = Cl- + 2OH- | 2 | | 2 |;

Mn2+ + 2(ClO)- + 4OH- = (MnO4)2- + 2Cl- + 2H2O.

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