Question

Se considera 112 l/mol Cl2 aflati in conditii normale.
a)Calculati:1)nr de moli.
2)masa(in grame)
3)nr de molecule

b)Notati ecuatia reactiei chimice cu hidrogenul

c)Calculati volumul de acid clorhidric rezultat masurat la 1,23 atmosfere si 17°C

d)Calculati densitatea absoluta a acidului clorhidric si densitatea fata de aer.


Va rog frumos sa ma ajutati!!!!E urgent!!!!Dau coroana...

Answer 1

niu.V/Vm= 112 l/22,4 l/mol= 5molCl2

m=niuxM= 5molx71g/mol=................................................calculeaza !!!!!!

N= niuxN,A= 5molx6,023x10²³molec/mol=30x10²³molec Cl2

1mol..............2mol

Cl2  +  H2 = 2HCl

5mol.................x=10mol

pV=niuRT        1,23xV=10x0,082x(273+17)-->-V=...................calculeaza !!!

ro=M/Vm= 71g/mol/22,4 l/mol=3,16g/l

M,aer= 29g/mol

d=M/M,aer= 71/29=.....................................................................calculeaza !!!!!

Answer 2

VCl2=112  litri

n=nr. moli

n=112 litri  : 22,4 litri/moli=5 moli clor

masa =m

m=n .M       ;   MCl2=2.35,5=71-------> 71 g/moli

m=5moli . 71 g/moli=355 g clor

1mol clor----------6,023 .10²³ molecule

5moli -------------x

x=5 . 6,023 .10²³ molecule =30,115 .10²³ molecule

b.

   5moli             ymoli

      Cl2 + H2=2HCl

    1mol             2moli

y=5.2=10 moli HCl

c.

pV = n R T

V=n. R  .T : p

V=10 moli .0,082 litri . atm /mol . K  . (17 +273) : 1,23 atm

V=193,3 litri HCl