дано
m(CH3COOH) = 60 g
η(CH3COOC2H5) - 90%
mпр.(CH3COOC2H5)-?
CH3COOH+C2H5OH-->CH3COOC2H5+H2O
M(CH3COOH) = 60 g/mol
n(CH3COOH) = m/M = 60 / 60 = 1 mol
n(CH3COOH) = n(CH3COOC2H5) = 1 mol
M(CH3COOC2H5) = 88 g/mol
m(CH3COOC2H5) = n*M = 1*88 = 88 g
m пр.(CH3COOC2H5) = 88 * 90% / 100% = 79.2 g
ответ 79.2 г
Объяснение:
дано
m(CH3COOH) = 60 g
η(CH3COOC2H5) - 90%
mпр.(CH3COOC2H5)-?
CH3COOH+C2H5OH-->CH3COOC2H5+H2O
M(CH3COOH) = 60 g/mol
n(CH3COOH) = m/M = 60 / 60 = 1 mol
n(CH3COOH) = n(CH3COOC2H5) = 1 mol
M(CH3COOC2H5) = 88 g/mol
m(CH3COOC2H5) = n*M = 1*88 = 88 g
m пр.(CH3COOC2H5) = 88 * 90% / 100% = 79.2 g
ответ 79.2 г
Объяснение: