Bonjour,
1) C₄H₆O + C₁₀H₁₈O = C₁₂H₂₀O₂ + C₂H₄O₂
2) M(C₄H₆O) = 4x12 + 6x1 + 16 = 70 g.mol⁻¹ (on va noter M(A))
M(C₁₀H₁₈O) = 10x12 + 18x1 + 16 = 154 g.mol⁻¹ (on va noter M(L))
ρA = 1,08 g.mL⁻¹ ⇒ m(A) = V x ρA = 10,0 x 1,08 = 10,8 g
⇒ nA = m(A)/M(A) = 10,8/70 = 0,154 mol
ρL = 0,87 g.mol⁻¹ ⇒ m(L) = V' x ρL = 5,0 x 0,87 = 4,35 g
⇒ n(L) = m(L)/M(L) = 4,35/154 = 0,028 mol
3)
C₄H₆O + C₁₀H₁₈O = C₁₂H₂₀O₂ + C₂H₄O₂
Etat avanct x
initial 0 n(A) n(L) 0 0
en cours x n(A) - x n(L) - x x x
final xf n(A) - xf n(L) - xf xf xf
4)
n(A) - xf = 0 ⇒ xf = n(A) = 0,154 mol
n(L) - xf = 0 ⇒ xf = n(L) = 0,028 mol
⇒ xmax = 0,028 mol
⇒ le réactif limitant est l'alcool (linalol)
5) n(E) = xmax = 0,028 mol
6) M(E) = 12x12 + 20x1 + 2x16 = 196 g.mol⁻¹
⇒ m(E) = n(E) x M(E) = 0,028 x 196 = 5,49 g
ρE = 0,89 g.mL⁻¹
⇒ V(E) = m(E)/ρE = 5,49/0,89 = 6,17 mL
