дано
m(HCOOH) = 46 g
η(HCOOC2H5) = 80%
mпр.(HCOOC2H5)-?
HCOOH+C2H5OH -->HCOOC2H5+H2O
M(HCOOH) = 46 g/mol
n(HCOOH) = m/M = 46 / 46 = 1 mol
n(HCOOH) = n(HCOOC2H5) = 1 mol
M(HCOOC2H5) = 74 g/mol
m(HCOOC2H5) = n*M = 1 * 74 = 74 g
mпр.(HCOOC2H5) = 74 * 80% / 100% = 59.2 g
ответ 59.2 г
Объяснение:
дано
m(HCOOH) = 46 g
η(HCOOC2H5) = 80%
mпр.(HCOOC2H5)-?
HCOOH+C2H5OH -->HCOOC2H5+H2O
M(HCOOH) = 46 g/mol
n(HCOOH) = m/M = 46 / 46 = 1 mol
n(HCOOH) = n(HCOOC2H5) = 1 mol
M(HCOOC2H5) = 74 g/mol
m(HCOOC2H5) = n*M = 1 * 74 = 74 g
mпр.(HCOOC2H5) = 74 * 80% / 100% = 59.2 g
ответ 59.2 г
Объяснение: