" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
MCH3OH=12+3+16+1=32g/mol
MNa=23g/mol
MCH3ONa=54g/mol
jest nadmiar CH3OH czyli sód cał przereaguje
108g-46gNa
xg-2,3g
xg=5,4
1molCH3ONa-54g
n moli-5,4g
n=0,1mola
czyli na kazde 0,1 mola metanolanu sodu przypada 0,1 metanolu a wiec na 1 mol CH3ONa przypdada 1 mol CH3OH