Chọn đáp án B
Ta có:
$\{\begin{array}{l}{n}_{\text{NaOH}}=0,1\to n_{\text{COO}}=0,1\\ \text{RCOONa}\stackrel{\text{chảy}}{\to }{n}_{{\text{H}}_2\text{O}}=0,05\end{array}$ $\to$ $n_{\text{HCOONa}}$ = 0,1

$\stackrel{\text{BTKL}}{\to }$ 6,18 + 0,1.40 = 0,1.68 + 3,2 + 18n${}_{{\text{H}}_2\text{O}}$
$\to$ n${}_{{\text{H}}_2\text{O}}$ = 0,01
$\stackrel{\text{BT.COO}}{\to }$ n${}_{\text{este}}$ = 0,1 - 0,01 = 0,09
$\to$ $\%{\text{HCOOCH}}_3=\frac{0,09.60}{6.18}$ = 87,38\%