For the reaction 2KClO3–>2KCl+3O2 how many Moles of potassium chlorate are required to produce 96 g of oxygen?
回答 (2)
2017-12-18 3:57 pm
Molar mass of O₂ = 16.0 × 2 = 32.0 g/mol
No. of moles of O₂ produced = (96 g) / (32.0 g/mol) = 3 mol
2KClO₃ → 2KCl + 3O₂
Mole ratio KClO₃ : O₂ = 2 : 3
No. of moles of KClO₃ required = (3 mol) × (2/3) = 2 mol
No. of moles of O₂ produced = (96 g) / (32.0 g/mol) = 3 mol
2KClO₃ → 2KCl + 3O₂
Mole ratio KClO₃ : O₂ = 2 : 3
No. of moles of KClO₃ required = (3 mol) × (2/3) = 2 mol
2017-12-18 12:31 pm
2 moles KClO3
收錄日期: 2021-04-18 18:01:21
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