Question

Giúp e bài 14 chi tiết đi ạ

Answer 1

\(n_{CH_3COOC_2H_5}=\dfrac{12,3}{88}=\dfrac{123}{880}\left(mol\right)\)

PTHH: CH₃COOH + C₂H₅OH --H₂SO_4(đ),t^o--> CH₃COOC₂H₅ + H₂O

                  \(\dfrac{123}{880}\)<----------------------------------\(\dfrac{123}{880}\)

=> \(\%m_{CH_3COOH\left(pư\right)}=\dfrac{\dfrac{123}{880}.60}{12}.100\%=69,89\%\)