Chlorine gas reacts with fluorine gas to form chlorine trifluoride.
Cl2(g)+3F2(g)→2ClF3(g)
A 2.30 L reaction vessel, initially at 298 K, contains chlorine gas at a partial pressure of 337 mmHg and fluorine gas at a partial pressure of 856 mmHg .
Identify the limiting reactant and determine the theoretical yield of ClF3 in grams.
Chemistry question HELLLLPP!!!?
2017-11-05 3:31 pm
回答 (1)
2017-11-05 4:21 pm
Molar mass of ClF₃ = (35.5 + 19.0×3) g/mol = 92.5 g/mol
(The molar mass learned in lessons should be used instead.)
Cl₂(g)+3F₂(g)→2ClF₃(g)
Mole ratio Cl₂ : F₂ : ClF₃ = 1 : 3 : 2
At constant volume and temperature, mole ratio of gases is equal to the pressure ratio.
When F₂ gas at a partial pressure of completely reacts :
Partial pressure of Cl₂ needed = (856 mmHg) × (1/3) = 285.3 mmHg < 337 mmHg
Hence, Cl₂ is in excess, and thus F₂ is the limiting reactant.
Partial pressure of ClF₃ formed = (856 mmHg) × (2/3) = 570.7 mmHg = 570.7/760 atm
Gas law : PV = nRT
Then, n = PV/(RT)
Maximum no. of moles of ClF₃ formed, n = (570.7/760) × 2.30 / (0.08206 × 298) mol = 0.07063 mol
Theoretical yield of ClF₃ = (0.07063 mol) × (92.5 g/mol) = 6.53 g
(The molar mass learned in lessons should be used instead.)
Cl₂(g)+3F₂(g)→2ClF₃(g)
Mole ratio Cl₂ : F₂ : ClF₃ = 1 : 3 : 2
At constant volume and temperature, mole ratio of gases is equal to the pressure ratio.
When F₂ gas at a partial pressure of completely reacts :
Partial pressure of Cl₂ needed = (856 mmHg) × (1/3) = 285.3 mmHg < 337 mmHg
Hence, Cl₂ is in excess, and thus F₂ is the limiting reactant.
Partial pressure of ClF₃ formed = (856 mmHg) × (2/3) = 570.7 mmHg = 570.7/760 atm
Gas law : PV = nRT
Then, n = PV/(RT)
Maximum no. of moles of ClF₃ formed, n = (570.7/760) × 2.30 / (0.08206 × 298) mol = 0.07063 mol
Theoretical yield of ClF₃ = (0.07063 mol) × (92.5 g/mol) = 6.53 g
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