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

Moles of `CH_(3)COOH` in 0.1 M solution having volume 100 mL `= (0.1)/(1000) xx 100 = 0.01` moles `CH_(3)COOH`.
`CH_(3)COOH + NaOH rarr CH_(3) COONa + H_(2)O`
1 mol of `CH_(3) COOH` gives 1 mol of `CH_(3)COONa` 0.01 mol of `CH_(3)COOH` gives 0.01 mol of `CH_(3)COONa`
`2CH_(3)COONa + H_(2)O rarr CH_(3) - CH_(3) + 2CO_(2) + NaOH + H_(2)`
2 moles of `CH_(3)COONa` give 1 mol of ethane (22,400 `cm^(3)` at S.T.P.).
`therefore` 0.01 mol of `CH_(3)COONa` give ethane
`= (22,400)/(2) xx 0.01`
`= 112 cm^(3)`