Bonjour,
1)
M(C₇H₅O₃N) = 7xM(C) + 5xM(H) + 3xM(O) + M(N)
= 7x12,0 + 5x1,0 + 3x16,0 + 1x14,0
= 151,0 g.mol⁻¹
M(C₃H₆O) = ... = 58,0 g.mol⁻¹
2) n(C₇H₅O₃N) = m/M(C₇H₅O₃N)
= 1,00/151
≈ 6,62.10⁻³ mol
3) n(HO⁻) = Cs x V'
soit n(HO⁻) = 2,0 x 5,0.10⁻³ = 1,0.10⁻² mol
4) m(C₃H₆O) = ρ(C₃H₆O) x V
et n(C₃H₆O) = m(C₃H₆O)/M(C₃H₆O)
⇒ n(C₃H₆O) = ρ(C₃H₆O) x V/M(C₃H₆O)
ρ(C₃H₆O) = 1,05 g.mL⁻¹ = 1,05.10³ g.L⁻¹
soit n(C₃H₆O) = 1,05.10³ x 10,0.10⁻³/58,0 ≈ 1,81.10⁻¹ mol
5)
Equation 2C₇H₅O₃N + 2C₃H₆O + 2HO⁻ → C₁₆H₁₀N₂O₂ + 2C₂H₃O₂⁻ + 4H₂O
Etat Avct
initial 0 6,62.10⁻³ 1,81.10⁻¹ 1,0.10⁻² 0 0 0
Inter. x "" - 2x "" - 2x "" - 2x x 2x 4x
Final xmax "" -2xmax etc...
Détermination de xmax :
6,62.10⁻³ - 2xmax = 0 ⇒ xmax = 3,31.10⁻³ mol
1,81.10⁻¹ - 2xmax = 0 ⇒ xmax = 9,05.10⁻² mol
Donc xmax = 3,31.10⁻³ mol
6) Le réactif limitant est le C₇H₅O₃N
7) nmax(indigo) = xmax = 3,31.10⁻³ mol
8) M(indigo) = 16x12,0 + 10x1,0 + 2x14,0 + 2x16,0 = 262,0 g.mol⁻¹
⇒ m(indigo) = nmax(indigo) x M(indigo)
soit m(indigo) = 3,31.10⁻³ x 262,0 ≈ 8,67.10⁻¹ g
