Cho 3,15 gam hỗn hợp X gồm axit axetic, axit acrylic, axit propionic vừa đủ để

${n_{B{r_2}}} = 0,02{\rm{ mol}}$
Ta có : $\left\{ \begin{array}{l} {m_{C{H_3}COOH}} + {m_{C{H_2}CHCOOH}} + {m_{C{H_3}C{H_2}COOH}} = 3,15\\ {n_{C{H_2}CHCOOH}} = {n_{B{r_2}}} = 0,02{\rm{ mol}}\\ {n_{C{H_3}COOH}} + {n_{C{H_2}CHCOOH}} + {n_{C{H_3}C{H_2}COOH}} = {n_{NaOH}} = 0,045{\rm{ mol}} \end{array} \right.$
$\to \left\{ \begin{array}{l} 60{n_{C{H_3}COOH}} + 72{n_{C{H_2}CHCOOH}} + 74{n_{C{H_3}C{H_2}COO}} = 3,15\\ {n_{C{H_2}CHCOOH}} = {n_{B{r_2}}} = 0,02{\rm{ mol }}\\ {n_{C{H_3}COOH}} + {n_{C{H_2}CHCOOH}} + {n_{C{H_3}C{H_2}COOH}} = 0,045{\rm{ mol}} \end{array} \right. \to \left\{ \begin{array}{l} {n_{C{H_3}COOH}} = 0,01\\ {n_{C{H_2}CHCOOH}} = 0,02\\ {n_{C{H_3}C{H_2}COOH}} = 0,015 \end{array} \right.$
$\to {m_{C{H_{_3}}COOH}} = 0,6{\rm{ gam}} \to {\rm{\% C}}{{\rm{H}}_{\rm{3}}}{\rm{COO}}{{\rm{H}}_{\left( x \right)}} = 19,05\%$