Question

помогите по химии(2)!20 баллов​

Answer 1

Ответ:

дано

m(CH3COOH) = 28,8 g

m(CH3OH) = 19.2 g

m практ(CH3COOCH3) = 29.6 g

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η(CH3COOCH3) - ?

M(CH3COOH) = 60 g/mol

n(CH3COOH) = m(CH3COOH) / M(CH3COOH) = 28.8 / 60 = 0.48 mol

M(CH3OH) = 32 g/mol

n(CH3OH) = m(CH3OH) / M(CH3OH)  = 19.2 / 32 = 0.6 mol

n(CH3COOH) < n(CH3OH)

CH3COOH+CH3OH-->CH3COOCH3+H2O

n(CH3COOH) = n(CH3COOCH3) = 0.48 mol

Находим массу теоретическую сл.эфира по УХР

M(CH3COOCH3) = 74 g/mol

m(CH3COOCH3) = n(CH3COOCH3) * M(CH3COOCH3) = 0.48* 74 =35.52 g

η(CH3COOCH3) = m практ (CH3COOCH3) / m(CH3COOCH3) * 100%

n(CH3COOCH3) = 29.6 / 35.52 * 100% = 83.3%

ответ 83.3%

Объяснение: