B RCOOH + KOH  RCOOK + H2O RCOOH + NAOH  RCOONA + H2O NNAOH = NKOH...

Bài 5: B

RCOOH + KOH  RCOOK + H

2

O

RCOOH + NaOH  RCOONa + H

2

O

n

NaOH

= n

KOH

= 0,5.0,12 = 0,06 mol

ĐLBTKL: m

X

+ m

NaOH

+ m

KOH

= m

rắn

+

m

H

2

O

 m = 1,08 gam  n = 0,06 mol

H

2

O

H

2

O

60

,

3 = 60  R = 15

 n

RCOOH

= n = 0,06 mol  M

X

= R + 45 =

H

2

O

0

06

 X: CH

3

COOH  Đáp án B