How do you balance $Zn(OH)_2 + CH_3COOH -> Zn(CH_3COO)_2 + H_2O$ ?

Question

2642 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-23T23:15:17

Use the proton balance method given below to obtain:

$2"CH"_3"COOH" + "Zn"("OH")_2 rarr "Zn"("CH"_3"COO")_2 + 2"H"_2"O"$ .

One way to do it is to think of it as a proton donation (acid) reaction and a proton acceptance (base) reaction:

Acid:

$"CH"_3"COOH" rarr "CH"_3"COO"^{-} + "H"^+$

Base:

$"Zn"("OH")_2 + 2 "H"^{+} rarr "Zn"^{2+} + 2 "H"_2"O"$

The acid gives one mole of protons but tge base takes two. To balance the protons take two moles of the acid to one of the base:

$2"CH"_3"COOH" rarr 2"CH"_3"COO"^{-} +2 "H"^+$

$"Zn"("OH")_2 + 2 "H"^{+} rarr "Zn"^{2+} + 2 "H"_2"O"$

Then add them up. The protons cancel and the remaining ions combine to make the salt:

$2"CH"_3"COOH" + "Zn"("OH")_2 rarr "Zn"("CH"_3"COO")_2 + 2"H"_2"O"$

How do you balance Zn(OH)_2 + CH_3COOH -> Zn(CH_3COO)_2 + H_2OZn(OH)_2 + CH_3COOH -> Zn(CH_3COO)_2 + H_2O?