Question

KMnO4 + H2O2 + H2SO4 -> KI + H2O2 + H2SO4 -> Fe(OH)2 + H2O2 -> Hg(NO3)2 + H2O2 + NaOH -> Hg + ... K3 [Cr(OH)6] + H2O2 -> K2CrO4 + ...

Answer 1

MnO4 - +8H+ + 5e- = Mn2+ + 4H2O | x 2
H2O2 - 2e- = O2 + 2H+ | x 5
5H2O2 + 2MnO4 - +16H+ = 2Mn2+ + 5O2 + 10H+ + 8H2O
5H2O2 + 2MnO4 - +6H+ = 2Mn2+ + 5O2 + 8H2O
5H2O2 + 2K+ + 2MnO4 - +6H+ + 3SO4 2- = 2Mn2+ + 2K+ + 3SO4 - +5O2 + 8H2O
5H2O2 + 2KMnO4 + 3H2SO4 = 2MnSO4 + 5O2↑ + K2SO4 + 8H2O                           2KI + H2O2 + H2SO4 = K2SO4 + I2 + 2H2O                                                               2Fe(OH)2+H2O2(разб)= 2Fe(OH)3                                                                             2Hg(NO3)2 + H2O2 + 2NaOH = Hg2(NO3)2 + 2NaNO3 + 2H2O + O2                             2 K3[Cr(OH)6] + 3 H2O2 -----> 2 K2CrO4 + 2 KOH + 8 H2O.

Answer 2

I can help you with your chemistry question. The user input seems to be a series of chemical reactions, but some of them are incomplete or incorrect. Here is a possible way to complete and correct them:

$$\text{KMnO}_4 + \text{H}_2\text{O}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{MnSO}_4 + \text{K}_2\text{SO}_4 + \text{O}_2 + \text{H}_2\text{O}$$

This is a redox reaction where $\text{MnO}_4^-$ is reduced to $\text{Mn}^{2+}$ and $\text{H}_2\text{O}_2$ is oxidized to $\text{O}_2$.

$$\text{KI} + \text{H}_2\text{O}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{K}_2\text{SO}_4 + \text{I}_2 + \text{H}_2\text{O}$$

This is another redox reaction where $\text{I}^-$ is oxidized to $\text{I}_2$ and $\text{H}_2\text{O}_2$ is reduced to $\text{H}_2\text{O}$.

$$\text{Fe(OH)}_2 + \text{H}_2\text{O}_2 \rightarrow \text{Fe(OH)}_3$$

This is a decomposition reaction where $\text{H}_2\text{O}_2$ breaks down into $\text{H}_2\text{O}$ and $\text{O}$, and the latter combines with $\text{Fe(OH)}_2$ to form $\text{Fe(OH)}_3$.

$$\text{Hg(NO}_3\text{)}_2 + \text{H}_2\text{O}_2 + \text{NaOH} \rightarrow \text{Hg} + \text{NaNO}_3 + \text{H}_2\text{O} + \text{O}_2$$

This is a complex reaction that involves both redox and precipitation. $\text{Hg}^{2+}$ is reduced to $\text{Hg}$, $\text{H}_2\text{O}_2$ is oxidized to $\text{O}_2$, and $\text{OH}^-$ precipitates $\text{Hg}^{2+}$ as $\text{HgO}$, which then decomposes into $\text{Hg}$ and $\text{O}_2$.

$$\text{K}_3[\text{Cr(OH)}_6] + \text{H}_2\text{O}_2 \rightarrow \text{K}_2\text{CrO}_4 + \text{KOH} + \text{H}_2\text{O}$$

This is a redox reaction where $\text{Cr}^{3+}$ is oxidized to $\text{CrO}_4^{2-}$ and $\text{H}_2\text{O}_2$ is reduced to $\text{H}_2\text{O}$.

If you want to learn more about how to balance chemical equations, you can check out this [tutorial](https://www.w3schools.com/python/python_user_input.asp) or this [interactive tool](https://stackoverflow.com/questions/37893927/python-how-to-take-user-input-and-use-that-in-function). You can also use Python to write a program that can balance equations for you. Here is an [example](https://realpython.com/python-input-output/) of how to do that. I hope this helps.