Question

O solutie de fier ( III ) se trateaza cu 6 moli de NaOH . Precipitatul format reactioneaza cu acidul sulfuric . Determinati masa de sare formata in final .

Answer 1

Fe + 3 NaOH ---> Fe(OH)3 + 3 Na
1 mol Fe(OH)3......3 moli NaOH
x moli Fe(OH)3.........6 moli NaOH      x = 2 moli
2 Fe(OH)3 + 3 H2SO4 ---> Fe2(SO4)3 + 6 H2O
2 moli Fe(OH)3.........1 mol Fe2(SO4)3
2 moli Fe(OH)3.............y moli Fe2(SO4)3            y = 1 mol sare
nr moli=masa/masa molara=>masa = 1 * 400 = 400 g sare
masa molara = 2*56+3*(32+4*16)=400g/mol