Ответ:
дано
m(CH3COOH) = 60 g
m( пр.CH3COOC2H5) = 80 g
------------------------------
η(CH3COOC2H5)-?
CH3COOH+C2H5OH-->CH3COOC2H5+H2O
M(CH3COOH) = 60 g/mol
n(CH3COOH) = m(CH3COOH) / M(CH3COOH) = 60 / 60 = 1 mol
n(CH3COOH) = n(CH3COOC2H5) = 1 mol
M(CH3COOC2H5) = 88 g/mol
m теор(CH3COOC2H5) = n(CH3COOC2H5) * M(CH3COOC2H5)
m теор(CH3COOC2H5) = 1*88 = 88 g
η(CH3COOC2H5) = m пр.(CH3COOC2H5) / m(теор.CH3COOC2H5) * 100%
η(CH3COOC2H5) = 60 / 88 * 100% = 68.18%
ответ 68.18%
Объяснение:
Copyright © 2024 SCHOLAR.TIPS - All rights reserved.
Answers & Comments
Ответ:
дано
m(CH3COOH) = 60 g
m( пр.CH3COOC2H5) = 80 g
------------------------------
η(CH3COOC2H5)-?
CH3COOH+C2H5OH-->CH3COOC2H5+H2O
M(CH3COOH) = 60 g/mol
n(CH3COOH) = m(CH3COOH) / M(CH3COOH) = 60 / 60 = 1 mol
n(CH3COOH) = n(CH3COOC2H5) = 1 mol
M(CH3COOC2H5) = 88 g/mol
m теор(CH3COOC2H5) = n(CH3COOC2H5) * M(CH3COOC2H5)
m теор(CH3COOC2H5) = 1*88 = 88 g
η(CH3COOC2H5) = m пр.(CH3COOC2H5) / m(теор.CH3COOC2H5) * 100%
η(CH3COOC2H5) = 60 / 88 * 100% = 68.18%
ответ 68.18%
Объяснение: