n1 moli C6H5-CH=O + CH3-CO-CH3 ⇒ C6H5-CH=CH-CO-CH3 + H2O
n2 moli 2C6H5-CH=O +CH3COCH3 ⇒ C6H5-CH=CHCOCH=CH-C6H5 +2H2O
M benzilidenactona = M1 = 146g/mol
M dibenzilidenacetona = M2 = 234g/mol
amestec final : m = 146n1 + 234·n2/2 = 146n1 + 117n2
146n1·100/m = 73 m = 200n1
117n2·100/m = 27 13n2·100/(200n1) = 3 n2 = 6n1/13
Cu = n1·100/(n1+n2) = 100n1/(19n1/13) = 68,42%