RÄspuns :
1. benzen : CâHâ (nâ moli) xilen : CâHââ (nâ moli)
% = (6n1 + 8n2)¡12/ (78n1 + 106n2) ¡100
91,76(78n1 + 106n2) = 100(6n1 + 8n2)¡12
7157,28n1 + 9726,56n2 = 7200n1 + 9600n2
126,56n2 = 42,72n1 n1/n2 = 126,56/42,72 = 2,96
2. CâHâ + HOSOâH = CâHâ SOâH + HâO Mâ = 158g/mol
CâHâ + 2HOSOâH = CâHâ(SOâH)â + 2HâO Mâ = 238g/mol
nâ¡Mâ + nâ¡Mâ = 356 n1/n2 = 3/1 n1 = 3n2
158¡3¡n2 + 238¡n2 = 356 712n2 = 356 n2 = 0,5kmoli n1 = 1,5kmoli
n benzen = n1 + n2 = 2kmoli m benzen = 156 kg
b) nH2SO4 = n1 + 2n2 = 1,5 + 1 = 2,5 kmoli mH2SO4 reactionat = 2,5¡98 =245kg
mH2SO4 luat in lucru = 245 + 30% ¡245 = 318,5 = md initial â
ms initial = 318,5¡100/98 = 325kg (318,5 kg acid + 6,5 kg apa)
acidul rezidual contine : (318,5 - 245 = 73,5 kg acid) si (6,5 + 2,5¡18 = 51,5kg apa)
â ms = 125kg
se adauga m kg oleum ( 0,8m kg acid + 0,2m kg SO3) SO3 + HOH = H2SO4
final : msf = 125 +m c = 98% md = 73,5 + 0,8m + 0,2/80 ¡98 ¡m = 73,5 + 1,045m
(73,5+1,045m)/(125+m) ¡100 = 98 7350 + 104,5m = 12250 + 98m
6,5m = 4900 m = 753,85 kg
% = (6n1 + 8n2)¡12/ (78n1 + 106n2) ¡100
91,76(78n1 + 106n2) = 100(6n1 + 8n2)¡12
7157,28n1 + 9726,56n2 = 7200n1 + 9600n2
126,56n2 = 42,72n1 n1/n2 = 126,56/42,72 = 2,96
2. CâHâ + HOSOâH = CâHâ SOâH + HâO Mâ = 158g/mol
CâHâ + 2HOSOâH = CâHâ(SOâH)â + 2HâO Mâ = 238g/mol
nâ¡Mâ + nâ¡Mâ = 356 n1/n2 = 3/1 n1 = 3n2
158¡3¡n2 + 238¡n2 = 356 712n2 = 356 n2 = 0,5kmoli n1 = 1,5kmoli
n benzen = n1 + n2 = 2kmoli m benzen = 156 kg
b) nH2SO4 = n1 + 2n2 = 1,5 + 1 = 2,5 kmoli mH2SO4 reactionat = 2,5¡98 =245kg
mH2SO4 luat in lucru = 245 + 30% ¡245 = 318,5 = md initial â
ms initial = 318,5¡100/98 = 325kg (318,5 kg acid + 6,5 kg apa)
acidul rezidual contine : (318,5 - 245 = 73,5 kg acid) si (6,5 + 2,5¡18 = 51,5kg apa)
â ms = 125kg
se adauga m kg oleum ( 0,8m kg acid + 0,2m kg SO3) SO3 + HOH = H2SO4
final : msf = 125 +m c = 98% md = 73,5 + 0,8m + 0,2/80 ¡98 ¡m = 73,5 + 1,045m
(73,5+1,045m)/(125+m) ¡100 = 98 7350 + 104,5m = 12250 + 98m
6,5m = 4900 m = 753,85 kg