Ответ:

дано

m(CH3COOH) = 120 g

+C2H5OH

η(CH3COOC2H5) = 90%

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m практ (CH3COOC2H5) = ?

CH3COOH+C2H5OH-->CH3COOC2H5+H2O

M(CH3COOH) = 60 g/mol

n(CH3COOH) = m(CH3COOH) / M(CH3COOH) = 120 / 60 = 2 mol

n(CH3COOH) = n(CH3COOC2H5) = 2 mol

M(CH3COOC2H5) = 88 g/mol

m теор(CH3COOC2H5) = n(CH3COOC2H5) * M(CH3COOC2H5)

m теор(CH3COOC2H5)= 2*88 = 176 g

m практ (CH3COOC2H5) =  m теор(CH3COOC2H5) * η(CH3COOC2H5) / 100%

m практ (CH3COOC2H5) = 176*90% / 100% = 158.4 г

ответ 158.4 гр

Решал: Овчинников Алексей Володимирович

Объяснение: