$X+3H_2O\rightleftharpoons 3$ axit béo $+$ glixerol.
$n_{X\text{pứ}}=0,15.60\%=0,09 mol$
$\to n_{RCOOH}=0,09.3=0,27 mol$
$M_{RCOOH}=\dfrac{69,12}{0,27}=256=R+45$
$\Leftrightarrow R=211 (C_{15}H_{31}-)$
Axit là axit panmitic.
Vậy X là tripanmitin.
$\to A$