Дано
m(CH3COOH) = 120 g
η = 90%
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m практ (CH3COOC2H5)-?
C2H5OH+CH3COOH-->CH3COOC2H5 + Н2О
M(CH3COOH) = 60 g/mol
n(CH3COOH) = m/M = 120 /60 = 2mol
n(CH3COOH) = n(CH3COOC2H5) = 2 mol
M(CH3COOC2H5) = 88 g/mol
m теор (CH3COOC2H5) = n*M = 2*88 = 176 g
m практ (CH3COOC2H5) = 176 *90% / 100% = 158.4 g
ответ 158.4 г