a. \(pthh:CH_3COOH+C_2H_5OH\xrightarrow[< ---]{H_2SO_{4_đ},t^o}CH_3COOC_2H_5+H_2O\)
Ta có: \(\left\{{}\begin{matrix}n_{CH_3COOH}=\dfrac{60}{60}=1\left(mol\right)\\n_{C_2H_5OH}=\dfrac{96}{46}=\dfrac{48}{23}\left(mol\right)\end{matrix}\right.\)
Ta thấy:\(\dfrac{1}{1}< \dfrac{\dfrac{48}{23}}{1}\)
Vậy rượu etylic dư
b. Theo pt: \(n_{CH_3COOC_2H_5}=n_{CH_3COOH}=1\left(mol\right)\)
\(\Rightarrow m_{CH_3COOC_2H_5}=1.88=88\left(g\right)\)
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