Question

Prin sulfonarea unei probe de benzen se obtine 3960g de amestec de acid benzensulfonic, acid benzendisulfonic si acid benzentrisulfonic in raport molar 6:3:1. A) masa solutiei h2so4 de concentratie 98% consumata, stiind ca acidul sulfuric se afla in exces de 20%

Answer 1

C6H6 + H2SO4------>C6H5-SO3H  +H2O   (1)

C6H6 +2 H2SO4------>C6H4(SO3H)2  +2H2O  (2)

C6H6 + 3H2SO4------>C6H3(SO3H)3  +3H2O (3) 

a)

x:y:z=6:3:1

158x+238y+318z=3960

x/y=6/3

y/z=3/1

se rezolva sistemul si x=12moli, y= 6moli, z=2 moli

12 moli ac. benzensulfonic

6 moli ac. benzen disulfonic

2 moli ac. benzen trisulfonic

din ec. reactiilor rezulta ca se consuma : 12+6+2=20 moli benzen

m C6H6=20*78=1560 g

b)

nr. moli H2SO4=30 moli (12 moli in prima reactie, 12 moli in a doua reactie si 6 moli in a treia )  

m h2SO4=30*98=2940g

ms= 2940*100/98=3000g     3000-2940=60 g H2O

3000* 20/100=600g excesul

ms=3000+600=3600g

c)

acidul sulfuric rezidual 600g contine 588g H2SO4 si 12g apa la care se adauga apa din sol. de ac. sulfuric consumat in reactii , adica 60 g DECi m ac. rezidual este 600+60=660g care contine [588g H2SO4 si 72 g apa]

x g oleum contin 80x/100 g H2SO4 si 20x/100 g SO3

80                 18                       98

SO3    +        H2O    ------> H2SO4

20x/100         a                      b   ==>    a=360x/8000 =36x/800   

mH2O ramasa=72-36x/800

                                               b=196x/800  (H2SO4)

x g oleum + 660g sol H2SO4    ==> o sol. ce contine  588g H2SO4

                                                                                 80x/100 (din oleum)

                                                                          196x/800 g H2SO4 (din reactie)

                                                                             ( 72- 36x/800) g H2O

md H2SO4=588+80x/100+196x/100

se inlocuiesc datele in relatia 98/100=md/ms de unde se afla valoarea lui x=.....