Question

REPEDE!!! VA ROG!!! 45 DE PUNCTE!
O cantitate de 16,73 g HCl impurificat cu clor formeaza in reactia cu fierul 28,65 g de saruri.
Calculati: a) compozitia procentuala a amestecului gazos initial; b) raportul molar al sarurilor obtinute.

Answer 1

n1 moli       2HCl + Fe ⇒ FeCl2 + H2
n2 moli       3Cl2 +2Fe ⇒2FeCl3
36,5n1 + 71n2 = 16,73                     36,5n1 + 71n2 = 16,73  |·(-108,33)
127n1/2 + 2/3 ·162,5n2 = 28,65      63,5n1 + 108,33n2 = 28,65 |·71
                                                        ----------------------------------------
                                                         4508,5n1- 3954n1 = 2034,15 - 1812,36
 554,5n1 = 221,79    n1 = 0 4 moli HCl     m1 = 0,4·36,5 = 14,6 g
m2 = 2,13gCl2    n2 = 0,03 moli Cl2
a) %HCl = 14,6·100/16,73 = 89,57%   %Cl2 = 213/16,73 = 12,73%
b) (n1/2) / (2n2/3) = 3n1/(4n2) = 1,2/0,12 = 10/1