Xà phòng hoá hoàn toàn 0,1 mol một este

Chọn đáp án C

Giải thích:

$\begin{array}{l}+\left\{\begin{array}{l}{n}_{MOH}=\frac{\mathrm{7,28}}{M+17}\\ n_{M_{2}C{O}_3}=\frac{\mathrm{8,97}}{2M+60}\end{array}\right.;\left\{\begin{array}{l}BTM:n_{MOH}=2n_{M_{2}C{O}_3}\\ hay\frac{\mathrm{7,28}}{M+17}=\frac{\mathrm{2.8,97}}{2M+60}\end{array}\right.\Rightarrow \left\{\begin{array}{l}M=39\left(K\right)\\ n_{KOH}=\mathrm{0,13}\\ n_{K_{2}C{O}_3}=\mathrm{0,065}\end{array}\right.\\ \Rightarrow Ygom\left\{\begin{array}{l}\underset{\mathrm{0,1}mol}{\underbrace{RCOOK}}\\ \underset{\mathrm{0,03}mol}{\underbrace{KOHdö}}\end{array}\right.\Rightarrow m_Y=\mathrm{0,1.}(R+83)+\mathrm{0,03.56}=\mathrm{12,88}\Rightarrow R=29(C_2{H}_5-).\\ +BTC:3\underset{\mathrm{0,1}}{\underbrace{n_{C_2{H}_{5}COOK}}}=\underset{?}{\underbrace{n_{C{O}_2}}}+\underset{\mathrm{0,065}}{\underbrace{n_{K_{2}C{O}_3}}}\Rightarrow n_{C{O}_2}=\mathrm{0,235}mol\Rightarrow V_{C{O}_2\text{\hspace{0.17em}\hspace{0.17em}d}ktc)}=\boxed{\mathrm{5,264}lít}\end{array}$