The volume of methane, ch4 at STP required to completely consume 3.50L of oxygen at STP?
回答 (2)
2016-07-05 8:32 am
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)
Volume ratio CH₄ : O₂ = 1 : 2
Volume of O₂ reacted = 3.5 L
Volume of CH₄ required = (3.5 L) × (1/2) = 1.75 L
Volume ratio CH₄ : O₂ = 1 : 2
Volume of O₂ reacted = 3.5 L
Volume of CH₄ required = (3.5 L) × (1/2) = 1.75 L
2016-07-05 8:34 am
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)
Volume ratio CH₄ : O₂ = 1 : 2
Volume of O₂ reacted = 3.5 L
Volume of CH₄ required = (3.5 L) × (1/2) = 1.75 L
Volume ratio CH₄ : O₂ = 1 : 2
Volume of O₂ reacted = 3.5 L
Volume of CH₄ required = (3.5 L) × (1/2) = 1.75 L
收錄日期: 2021-04-18 15:12:09
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