`B.` `65%`
`*`
`n_(CH_3COOH)=0,3` (mol)
`n_(CH_3OH)=0,2` (mol)
`*`
`CH_3COOH + CH_3OH` $\buildrel{{H_2SO_4\text{đặc}, t^o}}\over\rightleftharpoons$ `CH_3COOCH_3+H_2O`
`->n_(CH_3COOCH_3)=0,13` (mol)
`*` Hiệu suất phản ứng este hoá là
`H=(0,13)/(0,2).100=65%`