\(CH_3COOH+C_2H_5OH< =\left(H_2SO_4đ,t^o\right)=>CH_3COOC_2H_5+H_2O\)
\(n_{CH_3COOH}=\dfrac{60}{60}=1\left(mol\right)\)
Theo PTHH: \(n_{CH_3COOC_2H_5}\left(lt\right)=1\left(mol\right)\)
Vì \(H=90\%\)
\(\Rightarrow n_{CH_3COOC_2H_5}\left(tt\right)=\dfrac{1.90}{100}=0,9\left(mol\right)\)
Khối lượng Elyl Axetat thu được là:
\(\Rightarrow m_{CH_3COOC_2H_5}\left(tt\right)=0,9.88=79,2\left(g\right)\)
Đúng 0
Bình luận (4)
