Un volum de 0,5 L solutie de hidroxid de sodiu .NaOH,CU DENSITATEA DE 110 g\L si concentratie 10% reactioneaza cu clarura de fier trivalent.FeCl2.Ce masa de precipitat se va obtine. cine ma ajuta va rog.........

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FeCl3
3NaOH + FeCl3 -> 3NaCl + Fe(OH)3 ↓
m = ρ * v => masa de NaOH = 110 * 0.5 = 55g
c = md * 100/ms => md = c * ms/100
=> md = 10 * 55/100 = 5.5g
Masa molara NaOH = 23 + 16 + 1 = 40 g/mol
nr moli NaOH = 5,5/40 = 0.14 moli
Acum folosim regula de 3 simpla pentru a afla nr de moli de Fe(OH)3 :
3moli NaOH ... 1mol Fe(OH)3
0,14moli ...x => x = 0.14 * 1 / 3 = 0.05 moli
Masa molara Fe(OH)3=56+48 + 3 = 107g/mol
=> masa de Fe(OH)3 = 0.05 * 107 = 5.35 g
Sper ca te-am ajutat.
Nu am invatat regula de 3 simpla

Răspunsuri

2015-06-11T21:07:57+03:00
3NaOH + FeCl3 -> 3NaCl + Fe(OH)3 ↓
m = ρ * v => masa de NaOH = 110 * 0.5 = 55g
c = md * 100/ms => md = c * ms/100
=> md = 10 * 55/100 = 5.5g
Masa molara NaOH = 23 + 16 + 1 = 40 g/mol
nr moli NaOH = 5,5/40 = 0.14 moli
Acum folosim regula de 3 simpla pentru a afla nr de moli de Fe(OH)3 : 
3moli NaOH ... 1mol Fe(OH)3
0,14moli ...x => x = 0.14 * 1 / 3 = 0.05 moli
Masa molara Fe(OH)3=56+48 + 3 = 107g/mol
=> masa de Fe(OH)3 = 0.05 * 107 = 5.35 g
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