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2020-09-28 2:27 pm
Liquid hexane (CH3(CH2)CH3) will react with gaseous oxygen (O2)to produce gaseous carbon dioxide (CO2) and gaseous water (H2O) . Suppose 6.03 g of hexane is mixed with 29. g of oxygen. Calculate the Maximum mass of water that could be produced by the chemical reaction. Round to 3 significant digits.
回答 (2)
2020-09-28 2:58 pm
Molar mass of CH₃(CH₂)₄CH₃ = (12.0×6 + 1.0×14) g/mol = 86.0 g/mol
Molar mass of O₂ = 16.0×2 = 32.0 g/mol
Molar mass of H₂O = (1.0×2 + 16.0) g/mol = 18.0 g/mol
Initial moles of CH₃(CH₂)₄CH₃ = (6.03 g) / (86.0 g/mol) = 0.07012 mol
Initial moles of O₂ = (29 g) / (32.0 g/mol) = 0.9063 mol
2 CH₃(CH₂)₄CH₃(ℓ) + 19 O₂(g) → 12 CO₂(g) + 14 H₂O(g)
Mole ratio CH₃(CH₂)₄CH₃ :m O₂ = 2 : 19
If CH₃(CH₂)₄CH₃ completely reacts, O₂ needed = (0.07012 mol) × (19/2) = 0.6661 mol < 0.9063 mol
Hence, O₂ is in excess, and CH₃(CH₂)₄CH₃ is the limiting reactant.
According to the above equation, mole ratio CH₃(CH₂)₄CH₃ : H₂O = 2 : 14
Moles of H₂O produced = (0.07012 mol) × (14/2) = 0.4908 mol
Mass of H₂O produced = (0.4908 mol) × (18.0 g/mol) = 8.83 g
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OR:
If CH₃(CH₂)₄CH₃ is the limiting reactant:
(6.03 g CH₃(CH₂)₄CH₃) × (1 mol CH₃(CH₂)₄CH₃ / 86.0 g CH₃(CH₂)₄CH₃) × (14 mol H₂O produced / 2 mol CH₃(CH₂)₄CH₃) × (18.0 g H₂O / 1 mol H₂O)
= 8.83 g H₂O produced
If O₂ is the limiting reactant:
(29 g of O₂) × (1 mol O₂ / 32.0 g of O₂) × (14 mol H₂O produced / 19 mol O₂) × (18.0 H₂O / 1 mol H₂O)
= 12.0 g H₂O produced > 8.83 g H₂O
If completely reacted, the limiting reactant produced the smallest amount of produce.
Hence, CH₃(CH₂)₄CH₃ is the limiting reactant, and 8.83 g H₂O produced.
Molar mass of O₂ = 16.0×2 = 32.0 g/mol
Molar mass of H₂O = (1.0×2 + 16.0) g/mol = 18.0 g/mol
Initial moles of CH₃(CH₂)₄CH₃ = (6.03 g) / (86.0 g/mol) = 0.07012 mol
Initial moles of O₂ = (29 g) / (32.0 g/mol) = 0.9063 mol
2 CH₃(CH₂)₄CH₃(ℓ) + 19 O₂(g) → 12 CO₂(g) + 14 H₂O(g)
Mole ratio CH₃(CH₂)₄CH₃ :m O₂ = 2 : 19
If CH₃(CH₂)₄CH₃ completely reacts, O₂ needed = (0.07012 mol) × (19/2) = 0.6661 mol < 0.9063 mol
Hence, O₂ is in excess, and CH₃(CH₂)₄CH₃ is the limiting reactant.
According to the above equation, mole ratio CH₃(CH₂)₄CH₃ : H₂O = 2 : 14
Moles of H₂O produced = (0.07012 mol) × (14/2) = 0.4908 mol
Mass of H₂O produced = (0.4908 mol) × (18.0 g/mol) = 8.83 g
====
OR:
If CH₃(CH₂)₄CH₃ is the limiting reactant:
(6.03 g CH₃(CH₂)₄CH₃) × (1 mol CH₃(CH₂)₄CH₃ / 86.0 g CH₃(CH₂)₄CH₃) × (14 mol H₂O produced / 2 mol CH₃(CH₂)₄CH₃) × (18.0 g H₂O / 1 mol H₂O)
= 8.83 g H₂O produced
If O₂ is the limiting reactant:
(29 g of O₂) × (1 mol O₂ / 32.0 g of O₂) × (14 mol H₂O produced / 19 mol O₂) × (18.0 H₂O / 1 mol H₂O)
= 12.0 g H₂O produced > 8.83 g H₂O
If completely reacted, the limiting reactant produced the smallest amount of produce.
Hence, CH₃(CH₂)₄CH₃ is the limiting reactant, and 8.83 g H₂O produced.
2020-09-28 3:08 pm
The maximum mass of water produced is 8.83g
收錄日期: 2021-05-01 01:01:51
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