Question

Se supun descompunerii termice 2240L(c.n.) de CH4. Daca 25% din metan nu se transforma ,iar in amestecul final raportul molar CH4:H2=1:5 ,volumul (c.n.) de acetilena este...?

Answer 1

Reactia este 2CH4--->C2H2+ 3H2
nr. de moli=V/22,4 unde V=volumul
nr. de moli de metan=2240/22,4=100 moli din care 25% nu reactioneaza adica25/100*100=25 de moli nu reactioneaza
75 de moli de metan ii pui pe reactie si vine asa x(nr. de moli de C2H2)=1*75/2=75/2=37,5 moli de C2H2 
si conform primei formule 37,5=V/22,4 de unde V=22,4*37,5=840 de l in c.n.

Answer 2

2CH₄=C₂H₂+3H₂
Masa molara CH₄=1*12+4*1=16g/mol
Masa molara C₂H₂=2*12+2*1=26g/mol

1mol.......................22,4L CH₄
x moli.....................2240L CH₄
x=2240*1/22.4=100 moli CH₄

1mol.......................16g CH₄
100 moli...................y
y=100*16/1=1600g CH₄
1600*75/100=1200g CH₄ descompuse

z=masa de C₂H₂ obtinuta
z=1200*26/2*16=975g C₂H₂

1mol....................26g C₂H₂
q moli.................975g C₂H₂
q=975*1/26=37.5moli C₂H₂

1mol........................22.4L C₂H₂
37.5moli......................w
w=22.4*37.5/1=840L C₂H₂