\(n_{CH_3COOC_2H_5}=\dfrac{12,3}{88}=\dfrac{123}{880}\left(mol\right)\)
PTHH: CH3COOH + C2H5OH --H2SO4(đ),to--> CH3COOC2H5 + H2O
\(\dfrac{123}{880}\)<----------------------------------\(\dfrac{123}{880}\)
=> \(\%m_{CH_3COOH\left(pư\right)}=\dfrac{\dfrac{123}{880}.60}{12}.100\%=69,89\%\)