100 mL of 0.1 M acetic acid is completely n | vedclass

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(D) The neutralization reaction is: $CH_3COOH + NaOH ightarrow CH_3COONa + H_2O$. Moles of $CH_3COOH = 0.1 \ M 	imes 0.1 \ L = 0.01 \ mol$. Thus,$0

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$112$

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(D) The neutralization reaction is: $CH_3COOH + NaOH ightarrow CH_3COONa + H_2O$. Moles of $CH_3COOH = 0.1 \ M 	imes 0.1 \ L = 0.01 \ mol$. Thus,$0.01 \ mol$ of $CH_3COONa$ is formed. Kolbe's electrolysis reaction is: $2CH_3COONa + 2H_2O ightarrow C_2H_6 + 2CO_2 + H_2 + 2NaOH$. According to the stoichiometry,$2 \ mol$ of $CH_3COONa$ produces $1 \ mol$ of $C_2H_6$. Therefore,$0.01 \ mol$ of $CH_3COONa$ produces $0.005 \ mol$ of $C_2H_6$. Volume of $C_2H_6$ at $STP = 0.005 \ mol 	imes 22400 \ mL/mol = 112 \ mL$.

$100 \ mL$ of $0.1 \ M$ acetic acid is completely neutralized using a standard solution of $NaOH$. The volume of ethane obtained at $STP$ after the complete electrolysis of the resulting solution is (in $mL$)

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