\(n_{CH_3COOH}=\dfrac{4,8}{60}=0,08\left(mol\right)\)
\(V_{C_2H_5OH}=11,5.45\%=5,175\left(ml\right)\\ \rightarrow m_{C_2H_5OH}=5,175.0,8=4,14\left(g\right)\\ \rightarrow n_{C_2H_5OH}=\dfrac{4,14}{46}=0,09\left(mol\right)\)
\(PTHH:C_2H_5OH+CH_3COOH\underrightarrow{H_2SO_4đ,t^o}CH_3COOC_2H_5+H_2O\)
LTL: 0,09 > 0,08 => C2H5OH dư
Đề thiếu à :) ?