Bonjour,
1) 2H₂S + SO₂ → 3S + 2H₂O
2) m(H₂S) = 170,5 g
et M(H₂S) = 2xM(H) + M(S) = 2x1 + 32 = 34 g.mol⁻¹
⇒ ni(H₂) = m(H₂S)/M(H₂S) = 170,5/34 ≈ 5,0 mol
m(SO₂) = 192,3 g
et M(SO₂) = M(S) + 2xM(O) = 32 + 2x16 = 64 g.mol⁻¹
⇒ ni(SO₂) = m(SO₂)/M(SO₂) = 192,3/64 ≈ 3,0 mol
3)
2H₂S + SO₂ → 3S + 2H₂O
Etat Avanct.
initial 0 5,0 3,0 0 0
en cours x 5,0 - 2x 3,0 - x 3x 2x
final xf 5,0 - 2xf 3,0 - xf 3xf 2xf
4) 5,0 - 2xf = 0 ⇒ xf = 2,5 mol
ou 3,0 - xf = 0 ⇒ xf = 3,0 mol
⇒ xmax = 2,5 mol
Réactif limitant : H₂S
5) état final :
n(H₂S) = 0 mol
n(SO₂) = 3,0 - xmax = 0,5 mol
n(S) = 3xmax = 7,5 mol
n(H₂O) = 2xmax = 5 mol
6) m(S) = n(S) x M(S) = 7,5 x 32 = 240 g
