+3 -2 +2 0
FeCl3 + H2S --> FeCl2 + S + HCl
2x Fe⁺³__+e-_> Fe⁺² reducere (ag. Ox. FeCl3)
S⁻²__-2e-__> S⁰ oxidare (ag. Red. H2S)
2FeCl3 + H2S --> 2FeCl2 + S + 2HCl
5 moli n moli m g a g b moli
2FeCl3 + H2S --> 2FeCl2 + S + 2HCl
2 1 2x127 32 2
=> n = 1x5/2 = 2,5 moli H2S ar fi necesari => H2S in exces cu 0,5 moli
M.H2S= 34 g/mol => 34x0,5 = 17 g H2S in exces
=> m= 5x2x127/2 = 635 g FeCl2
=> a= 5x32/2 = 80 g S
=> b= 5x2/2 = 5 moli HCl
