How do you balance $Z n {(O H)}_{2} + C H_{3} C O O H \to Z n {(C H_{3} C O O)}_{2} + H_{2} O$ $Zn(OH)_2 + CH_3COOH -> Zn(CH_3COO)_2 + H_2O$ ?

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Answer 1 · 2016-08-23T23:15:17

Use the proton balance method given below to obtain:

$2 {CH}_{3} COOH + Zn {(OH)}_{2} \to Zn {({CH}_{3} COO)}_{2} + 2 H_{2} O$ $2"CH"_3"COOH" + "Zn"("OH")_2 rarr "Zn"("CH"_3"COO")_2 + 2"H"_2"O"$ .

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

Acid:

${CH}_{3} COOH \to {CH}_{3} {COO}^{-} + H^{+}$ $"CH"_3"COOH" rarr "CH"_3"COO"^{-} + "H"^+$

Base:

$Zn {(OH)}_{2} + 2 H^{+} \to {Zn}^{2 +} + 2 H_{2} O$ $"Zn"("OH")_2 + 2 "H"^{+} rarr "Zn"^{2+} + 2 "H"_2"O"$

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:

$2 {CH}_{3} COOH \to 2 {CH}_{3} {COO}^{-} + 2 H^{+}$ $2"CH"_3"COOH" rarr 2"CH"_3"COO"^{-} +2 "H"^+$

$Zn {(OH)}_{2} + 2 H^{+} \to {Zn}^{2 +} + 2 H_{2} O$ $"Zn"("OH")_2 + 2 "H"^{+} rarr "Zn"^{2+} + 2 "H"_2"O"$

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

$2 {CH}_{3} COOH + Zn {(OH)}_{2} \to Zn {({CH}_{3} COO)}_{2} + 2 H_{2} O$ $2"CH"_3"COOH" + "Zn"("OH")_2 rarr "Zn"("CH"_3"COO")_2 + 2"H"_2"O"$

How do you balance Zn(OH)_2 + CH_3COOH -> Zn(CH_3COO)_2 + H_2OZn(OH)2+CH3COOH→Zn(CH3COO)2+H2OZn(OH)_2 + CH_3COOH -> Zn(CH_3COO)_2 + H_2O?