Question

5. 400g fier de puritate 80% react cu o sol de h2so4 40% la un randament de 95%. se cere:
a) Vh2 degajat c.n.
b)Vh2 defajat mas. la 27grade si 3 atm
c) m sol h2so4 utilizata
d) h2 este amestecat cu 20g so3 si cu 80l N2 c.n
calculati:
e) M (cu linie deasupra) a amestecului
f) ro zero c.n
g)d in rap cu aerul
h) ce vol ocupa amestecul de mai sus la 17 grde si 2atm
i)ce v ocupa amestecul in c.n

Answer 1

0,8*400=320g Fe pur

56g.....98g.............152g.......2g
Fe + H2SO4 ---> FeSO4 + H2↑
320g...560g....868,571g....11,428g

a) n=m/M=11,428/2=5,714 moli H2
n=V/Vm==>V=n*Vm=5,714*22,4=128L H2

daca se ia randamentul in calcul => 5,714*95/100=5,4283 moli H2

V=22,4*5,4283=121,6L H2

b) pV=nRT==>V=nRT/p=5,714*0.082*300/3=46,8548L H2

cu randament => 5,4283*0.082*300/3=44,512L H2

c) Cp=md/ms*100==>ms=560*100/40=1400g H2SO4

d) ............

Answer 2

-calculez moli din fiecare gaz
n,H2=5,714mol
n,SO3= 20g/60g/mol= 0,333mol
n,N2= 80l/22,4l/mol= 3,571mol----.> SUMA = 9,618mol gaze

M, masa medie a amestrecului= X,H2x M,H2 + X,SO3xM,SO3 +X,N2XM,N3
X sunt fractii molare--> M= 5,714/ 9,618x2+0,333/9,618x60+3,571/9,618x28=
calculeaza !!!

ro,zero= M/ 22,4...calculeaza !!!!

d=M/M,aer= M/29.....calculeaza !!!!

2xV=9,618X0,082X(273+17)......calculeaza !!!!

V= 9,618molx22,4l/mol= calculeaza 

te rog sa verifici primele calcule !!!