\(a,n_{CH_3COOC_2H_5}=\dfrac{11}{88}=0,125\left(mol\right)\)
PTHH: \(CH_3COOH+C_2H_5OH\xrightarrow[t^o]{H_2SO_{4\left(đ\right)}}CH_3COOC_2H+H_2O\)
0,125<------------------------------0,125
b, => mCH3COOH = 0,125.60 = 7,5 (g)
\(c,H=\dfrac{7,5}{12}.100\%=62,5\%\)
a.b.\(n_{CH_3COOC_2H_5}=\dfrac{11}{88}=0,125mol\)
\(CH_3COOH+C_2H_5OH\rightarrow\left(t^o,H_2SO_4\left(đ\right)\right)CH_3COOC_2H_5+H_2O\)
0,125 0,125 ( mol )
\(m_{CH_3COOH}=0,125.60=7,5g\)
c.\(n_{CH_3COOH}=\dfrac{12}{60}=0,2mol\)
\(H=\dfrac{0,125}{0,2}.100\%=62,5\%\)