Question

4 K2Cr2O7 + KI + HCl = CrCl3 + I2 + KCl + H2O​

Answer 1

Ответ:

Перед KI ставь 6

Перед HCl ставь 14

Перед CrCl3 ставь 2

Перед I2 ставь 3

Перед KCl ставь 8

Перед H2O 7

Реакция ОВР, ну и собственно путем вычисления изменений степеней окисления можно расставить коэффициенты

Answer 2

The given chemical equation represents a reaction between potassium dichromate (K2Cr2O7), potassium iodide (KI), and hydrochloric acid (HCl) to produce chromium(III) chloride (CrCl3), iodine (I2), potassium chloride (KCl), and water (H2O). Here's the balanced equation for the reaction:

4 K2Cr2O7 + 16 KI + 28 HCl → 8 CrCl3 + 16 I2 + 16 KCl + 14 H2O

In this balanced equation, you can see that four molecules of potassium dichromate (K2Cr2O7) react with 16 molecules of potassium iodide (KI) and 28 molecules of hydrochloric acid (HCl). This results in the formation of eight molecules of chromium(III) chloride (CrCl3), 16 molecules of iodine (I2), 16 molecules of potassium chloride (KCl), and 14 molecules of water (H2O).