Bonjour pouvez vous m'aider s'il vous plaît merci.... Pergunta de ideia deVouriampa

January 2021 1 70 Report

Bonjour pouvez vous m'aider s'il vous plaît merci

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scoladan

Verified answer

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

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