How to balance this equation? [closed]








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Here is my problem


$$\ce{2X + 16H^{+} + 10Cl^{-} \rightarrow 2Mn^{2+} + 8H\_{2}O + 5Cl2}$$


I want to find the $X$ in this reaction. 


According to me


$$Mn\_{2}O\_{3}$$ 


Also I see that the values of $Cl\_2$ and $H$ are equal. There's $8$ Oxygen in subtitles as well $(8H\_{2}O)$.


Regards!




If these numbers are correct and there is just one compound X missing, it is quite simple:


You have 2 manganese on the right, so you need 2 on the left. 8 oxygen on the right, so 8 left. Because stoichiometry is 2, divide these numbers by 2.


This gives you $\ce{1Mn}$ and $\ce{4O}$, forming $\ce{MnO4}$.


**However**, we are still missing the charge!


If you look carefully, you see that on the right you have $2~\cdot~+2=+4$. On the left, you have $+16 -10=+6$. That is $+2$ too much, so you have to add 2 negative charges. Divide by stoichiometry of your compound $\ce{X}$ and add to what you have already.


So your compound $\ce{X}$ must be $\ce{MnO4-}$. Your equation is:


$$\ce{2MnO4- +16H+ + 10Cl- -> 2Mn^{2+} + 8H2O + 5Cl2}$$


This, however, is usually not the way to solve this kind of question.


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