Luam procentele ca fiind grame.
1mol....................120g
C6H6 + C3H6 -> C9H12
0,4033moli..........48,4g
1mol........................162g
C6H6 + 2C3H6 -> C12H18
0,0504moli..............8,167g
1mol........................204g
C6H6 + 3C3H6 -> C15H24
0,0201moli..............4,11g
1mol........78g
C6H6 -> C6H6
0,5041....39,322g
total: 0,978moli benzen
cantitatea trasformata=0,978-0,504=0,474moli
%trans=0,474*100%/0,978=48,46%