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PM H₂S = 2×(1g/mol×2 + 32g/mol×1) PM O₂ = 3× (16g/mol×2)
= 2× 34g/mol = 3× 32 g/mol
= 68 g = 96 g
2 H₂S + 3 O₂ → 2 H₂O + 2 SO₂
68 g H₂S → 96 g O₂
3 g H₂S → X
X = 3 g H₂S × 96 g O₂
68 g H₂S
X = 288 ⇒ 4,24 g O₂
68
Se necesitan 4,24 g O₂ para reaccionar con 3.00 g de H₂S (sulfuro de dihidrógeno)
porfa, calificame...gracias