Question

cum fac exerciitile 2 si 4?

Answer 1

La 2 este C
La 4 este E

Answer 2

Incepem prin scrierea ecuatiilor
2CH4 -> C2H2 + 3H2
3C2H2 -> C6H6
deoarece stim randamentul global al procesului, e mai comod la calcule sa folosim o singura reactie, astfel:
6CH4 -> C6H6 + 9H2
6CH4 ->9H2 + 3C2H2 (insa aceasta duce la benzen, de aceea putem scrie reactia direct)
m=d*V=0,88g/ml*100000ml=88000g=88kg benzen
randament/100=masapractica/masateoretica
masateoretica=masapractica*100/randament=88*100/70=125,71kg benzen
6*22,4m³|78kg
6CH4 -> C6H6 + 9H2
x=?.........125,71kg
x=216,6m³ CH4 => c)
4)notam raportul molar etil: dietil=x:y (pentru simplitate)
stim ca m=n*M
Metilbenzen=106kg/kmol
Mdietilbenzen=134kg/kmol
Mbenzen=78kg/kmol
raportul masic etil:dietil=106x:134y
78kg..............................106kg
C6H6 + CH2=CH2 -> C6H5-CH2-CH3
78x................................106x
78kg.................................134kg
C6H6 + 2CH2=CH2 -> C6H4(CH2CH3)2
78y...................................134y
avem sistem:
x+y=234/78=3 (am dat factor comun si am impartit) => x=3-y
106x+134y=346
106(3-y)+134y=318+28y=346 => 28y=28 => y=1mol
x=3-1=2moli
metilbenzen=2*106=212kg => e)