Bonjour,
1) Fe + 2Ag⁺ → Fe²⁺ + 2Ag
2)
ni(Fe) = m₁/M(Fe) = 0,41/55,8 ≈ 7,3.10⁻³ mol ou 7,3 mmol
ni(Ag) = C₂V₂ = 0,20 x 50,0.10⁻³ = 1,0.10⁻² mol ou 10 mmol
3)
Fe + 2Ag⁺ → Fe²⁺ + 2Ag
Etat Avanct
initial 0 7,3.10⁻³ 1,0.10⁻² 0 0
en cours x 7,3.10⁻³-x 1,0.10⁻²-2x x 2x
final xf 7,3.10⁻³-xf 1,0.10⁻²-2xf xf 2xf
4) Réactif limitant :
7,3.10⁻³ - xf = 0 ⇒ xf = 7,3.10⁻³ mol
ou
1,0.10⁻² - 2xf = 0 ⇒ xf = 5,0.10⁻³ mol
Donc : xf = 5,0.10⁻³ mol et le réactif limitant est Ag⁺
5) Etat final :
nf(Fe) = 7,3.10⁻³ - xf = 7,3.10⁻³ - 5,0.10⁻³ = 2,3.10⁻³ mol
nf(Ag⁺) = 0
nf(Fe²⁺) = xf = 5,0.10⁻³
nf(Ag) = 2xf = 1,0.10⁻² mol (soit 1,0.10⁻²x107,9 ≈ 1,08 g d'argent)
6) [Fe²⁺] = nf(Fe²⁺)/V₂ = 5,0.10⁻³/50,0.10⁻³ = 0,1 mol.L⁻¹
7) On veut m(Ag) = 2,5.10⁻³ g
Soit : n(Ag) = m(Ag)/M(Ag) = 2,5.10⁻³/107,9 ≈ 2,32.10⁻⁵ mol
a) il faut donc : n(Fe) = n(Ag)/2 ≈ 1,16.10⁻⁵ mol
soit : m(Fe) = n(Fe) x M(Fe) = 1,16.10⁻⁵ x 55,8 ≈ 6,46.10⁻⁴ g ≈ 0,65 mg
b) n(Ag⁺) = n(Ag)
⇒ V₂' = n(Ag⁺)/[Ag⁺] = 2,32.10⁻⁵/0,20 = 1,16.10⁻⁴ L ≈ 0,12 mL