Question

O proba cu masa de 15.6 g dintr-o hidrocarbura A ce are masa 78g/mol, se arde in 168 L aer(20%O2) procente volumetrice. O alta proba de hidrocarbura A se sulfoneaza cu o sol de acid sulfuric 98% formand un compus monosubstituit.Stiind ca intreaga cantitate de hidrocarbura A se consuma si ca numai 50% din cant de acid reactioneaza det: formula A si masa de oleu 20% SO3, ce trebuie adau in solutia finala de acid sulfuri pentru a o face reutilizabila.gata

Answer 1

CxH2y + (2x+y)/2 O2 = xCO2 + yH2O
n hidr. = 15,6/78 = 0,2 moli
nO2 = 0,1(2x+y) = 0,2·168/22,4 = 0,2·7,5      2x + y = 15             2x+ y = 15
                                                                            12x +2y = 78         6x + y = 39
                                                                                                          ----------------
                                                                                                            4x = 24    x = 6
A = C6H6
C6H6 + HOSO3H = C6H5-SO3H + H2O
initial : 0,2 moli C6H6 + msH2SO4(0,98m acid + 0,02m apa)
reactioneaza : 0,2 moli H2SO4 (0,2·98 = 19,6g) ⇒ 0,2 moli H2O
19,6g = 50% ·0,98m   ⇒  m = 40g solutie initiala de H2SO4 (39,2g H2SO4 + +0,8g apa)
solutia finala de H2SO4 : ms = 40 +3,6 - 19,6 = 24g    md = 39,2 - 19,6 = 19,6g
+ m (oleum)   (0,8mH2SO4 + 0,2mSO3)
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solutia reutilizabila : ms = 24 + m;  md = 19,6 + 0,8m + 0,2m/80 ·98 = 19,6+ +1,045m;  c = 98%
(19,6+ 1,045m)·100/(24+m) = 98
1960 + 104,5m = 2352 + 98m    6,5m = 392   m = 60,3 g