Oblicz ile gramow kwasu stearynowego potrzeba do reakcji z 14 gramami zasady potasowej
Reakcja
C17H35COOH + NaOH -----> C17H35COONa + H2O
! mol kwasu reaguje z 1 molem zasady
M kwasu = 18 x 12 + 36 x 1 + 2 x 16 = 284 g/mol
M NaOH = 23 + 17 = 40 g/mol
284 g C17H35COOH ------- 40 g NaOH
X -------------------------------- 14 g
X = 99,4 g kwasu stearynowego
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Reakcja
C17H35COOH + NaOH -----> C17H35COONa + H2O
! mol kwasu reaguje z 1 molem zasady
M kwasu = 18 x 12 + 36 x 1 + 2 x 16 = 284 g/mol
M NaOH = 23 + 17 = 40 g/mol
284 g C17H35COOH ------- 40 g NaOH
X -------------------------------- 14 g
X = 99,4 g kwasu stearynowego