Question

MnCl2+KBrO+KOH=MnO2+KBr​

Answer 1

Ответ:

надеюсь, тебе надо было что-то такое

Answer 2

The given chemical equation represents a reaction between manganese(II) chloride (MnCl2), potassium bromate (KBrO3), and potassium hydroxide (KOH) to form manganese(IV) oxide (MnO2) and potassium bromide (KBr).

The balanced chemical equation for the reaction is as follows:

3 MnCl2 + 6 KBrO3 + 6 KOH → 3 MnO2 + 6 KBr + 3 H2O

In this equation, you need three molecules of manganese(II) chloride (MnCl2), six molecules of potassium bromate (KBrO3), and six molecules of potassium hydroxide (KOH) to react. As a result, you obtain three molecules of manganese(IV) oxide (MnO2), six molecules of potassium bromide (KBr), and three molecules of water (H2O).