9,2 г этанол мен ..
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Ответ:

дано

m(C2H5OH) = 9.2g

m(CH3COOH) = 5.6 g

W(CH3COOC2H5) = 75%

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mпр (CH3COOC2H5) - ?

C2H5OH + CH3COOH-->CH3COOC2H5 + H2O

M(C2H5OH) = 46 g/mol

n(C2H5OH) = m/M = 9.2 / 46 = 0.2 mol

M(CH3COOH) = 60 g/mol

n(CH3COOH) = m/M = 5.6 / 60  = 0.09 mol

n(C2H5OH) > n(CH3COOH)

n(CH3COOH) = n(CH3COOC2H5)

M(CH3COOC2H5) = 88 g/mol

mтеор(CH3COOC2H5) = n*M = 0.09 * 88 = 7.92 g

mпр(CH3COOC2H5) = 7.92 * 75% / 100% = 5.94 g

ответ 5.94 г

Объяснение: