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

Options

A. 112 mL
B. 56 mL
C. 224 mL
D. 560 mL

Answer

A

Explanation

`CH_(3) CO OH + NaOH to CH_(3) CO O Na 2 CH_(3) CO O Na + 2 H_(2) O to CH_(3) - CH_(3) + 2 CO_(2) + 2 NaOH + H_(2)`
Number of moles of acetic acid = Number of moles of sodium acetate = `0.1 xx 100= 10` m moles = `10 xx 10^(-3)` moles
2 moles of `CH_(3) CO ONa` on electrolysis produces 22400 mL of ethane .
`therefore 10 xx 10^-3` moles of `CH_(3) CO ONa ` will produce = `(22400 xx 10 xx 10^(-3))/(2) = 112` mL of ethane