дано
m(CH3COH) = 66 g
η(CH3COOC2H5) = 75%
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m пр.(CH3COOC2H5)-?
5CH3-CHО+2KMnO4+3H2SO4 -->5CH3-COOH +2MnSO4+ K2SO4+3H2O
CH3COOH+C2H5OH-->CH3COOC2H5+H2O
M(CH3COH) = 44 g/mol
n(CH3COH) = m//M = 66 / 44 = 1.5 mol
5n(CH3COH) = n(CH3COOC2H5)
n(CH3COOC2H5) = 1.5 / 5 = 0.3 mol
M(CH3COOC2H5) = 88 g/mol
m теор.(CH3COOC2H5) = n*M = 0.3 * 88 = 26.4 g
m пр.(CH3COOC2H5) = 26.4 * 75% / 100% = 19.8 g
ответ 19.8 г