Question

Se oxideaza 40 g de toluen de puritate 92% si se obtin 40 de g de acid benzoic de puritate 73,2 %. care este randamentul reactiei? va rog mult, daca ma puteti ajuta

Answer 1

C6H5-CH3 +3[O]--->C6H5--COOH  +H2O
M toluen=92g/mol
M acid benzoic=122g/mol
m toluen pur=40 *92/100=36,8g
din ecuatie;
92g toluen..............122g acid benzoic
36,8g toluen..............x=48,8g acid benzoic   (mt)
practic se obtine 40g acid de p=73,2%
deci masa de acid puraeste 40*73,2/100=29,28 g (mp)
η=mp*100/mt =29,28*100/48,8= 60%