$RCOOH+NaOH\to RCOONa+H_2O$
Đặt $x$ là số mol $NaOH$
$\to n_{NaOH}=x(mol)$
BTKL:
$6+40x=8,2+18x$
$\to x=0,1(mol)$
$\to n_{RCOOH}=x=0,1(mol)$
$\to M_{RCOOH}=\dfrac{6}{0,1}=60$
$\to M_R=60-44-1=15(CH_3)$
Vậy CTCT $X$ là $CH_3COOH$ (axit axetic)
$\to$ CTPT: $C_2H_4O_2$