Question

Un amestec de clorura de argint si bromura de argint conține 60,94% argint in procente de masa. Determinați procentul de clor respectiv de brom din amestec.

Answer 1

AgCl  AgBr  60,94%Ag

x=%AgCl
y=%AgBr
a=%Cl
b=%Br

[tex]M _{AgCl =A_{Ag} +A_{Cl} =108+35,5=143,5 adica 1 mol AgCl =143,5 g deoarece a+b=100-60,94 = 39,06 M_{AgBr} =A_{Ag} +A_{Br} =108+80=188 adica 1 mol AgBr = 188 g 143,5 g AgCl ..............35,5 g Cl x g AgCl.............a g Cl unde x=143,5a / 35,5 188 g AgBr.......80 g Br y g AgBr ....b g Br unde y=188b/80[/tex]
$\left \{ {{x+y=100} \atop {a+b=39,06}} \right.$
$\left \{ {{x= \frac{143,5a}{35,5} } \atop {y= \frac{188b}{80} }} \right.$
[tex]143,5a/35,5+188b/80=100 unde a+b=39,06 rezulta ca a=4,85% Cl
 b=34,21% Br[/tex]