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A 10 L flask containing 10.8 g of N2O5 is heated to 373 K, which leads to its decomposition according to the equation 2 N2O5(g) → 4 NO2(g) + O2(g). If the final pressure in the flask is 0.5 atm, then the partial pressure of O2(g) in atm is __.
[Given R = 0.0821 L atm K–1 mol–1]
[Given R = 0.0821 L atm K–1 mol–1]
Correct answer is between '0.06,0.07'. Can you explain this answer?
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A 10 L flask containing 10.8 g of N2O5 is heated to 373 K, which leads...
Solution:
Given:
Initial volume V = 10 L
Initial mass of N2O5 = 10.8 g
Final pressure P = 0.5 atm
R = 0.0821 L atm K-1 mol-1
1. Finding the moles of N2O5:
The molar mass of N2O5 = 2 × 14.01 + 5 × 16 = 108.01 g/mol
Moles of N2O5 = Mass of N2O5 / Molar mass of N2O5
Moles of N2O5 = 10.8 g / 108.01 g/mol
Moles of N2O5 = 0.1 mol
2. Finding the moles of O2 produced:
As per the balanced chemical equation, 1 mole of N2O5 produces 0.5 moles of O2
Hence, Moles of O2 produced = 0.5 x Moles of N2O5
Moles of O2 produced = 0.5 x 0.1 mol
Moles of O2 produced = 0.05 mol
3. Finding the moles of NO2 produced:
As per the balanced chemical equation, 1 mole of N2O5 produces 2 moles of NO2
Hence, Moles of NO2 produced = 2 x Moles of N2O5
Moles of NO2 produced = 2 x 0.1 mol
Moles of NO2 produced = 0.2 mol
4. Finding the total moles of gases present:
Total moles of gases present = Moles of NO2 + Moles of O2
Total moles of gases present = 0.2 mol + 0.05 mol
Total moles of gases present = 0.25 mol
5. Finding the partial pressure of O2:
As per the given data, the final pressure in the flask is 0.5 atm.
Using the ideal gas equation PV = nRT, we can find the total pressure of the gases present in the flask.
Total pressure = (Total moles of gases present) x R x T / V
Total pressure = 0.25 mol x 0.0821 L atm K-1 mol-1 x 373 K / 10 L
Total pressure = 0.768 atm
The partial pressure of O2 can be found by multiplying the mole fraction of O2 with the total pressure.
Mole fraction of O2 = Moles of O2 / Total moles of gases present
Mole fraction of O2 = 0.05 mol / 0.25 mol
Mole fraction of O2 = 0.2
Partial pressure of O2 = Mole fraction of O2 x Total pressure
Partial pressure of O2 = 0.2 x 0.768 atm
Partial pressure of O2 = 0.1536 atm
Therefore, the partial pressure of O2 in atm is 0.1536 atm, which is between the given range of 0.06-0.07 atm.
Given:
Initial volume V = 10 L
Initial mass of N2O5 = 10.8 g
Final pressure P = 0.5 atm
R = 0.0821 L atm K-1 mol-1
1. Finding the moles of N2O5:
The molar mass of N2O5 = 2 × 14.01 + 5 × 16 = 108.01 g/mol
Moles of N2O5 = Mass of N2O5 / Molar mass of N2O5
Moles of N2O5 = 10.8 g / 108.01 g/mol
Moles of N2O5 = 0.1 mol
2. Finding the moles of O2 produced:
As per the balanced chemical equation, 1 mole of N2O5 produces 0.5 moles of O2
Hence, Moles of O2 produced = 0.5 x Moles of N2O5
Moles of O2 produced = 0.5 x 0.1 mol
Moles of O2 produced = 0.05 mol
3. Finding the moles of NO2 produced:
As per the balanced chemical equation, 1 mole of N2O5 produces 2 moles of NO2
Hence, Moles of NO2 produced = 2 x Moles of N2O5
Moles of NO2 produced = 2 x 0.1 mol
Moles of NO2 produced = 0.2 mol
4. Finding the total moles of gases present:
Total moles of gases present = Moles of NO2 + Moles of O2
Total moles of gases present = 0.2 mol + 0.05 mol
Total moles of gases present = 0.25 mol
5. Finding the partial pressure of O2:
As per the given data, the final pressure in the flask is 0.5 atm.
Using the ideal gas equation PV = nRT, we can find the total pressure of the gases present in the flask.
Total pressure = (Total moles of gases present) x R x T / V
Total pressure = 0.25 mol x 0.0821 L atm K-1 mol-1 x 373 K / 10 L
Total pressure = 0.768 atm
The partial pressure of O2 can be found by multiplying the mole fraction of O2 with the total pressure.
Mole fraction of O2 = Moles of O2 / Total moles of gases present
Mole fraction of O2 = 0.05 mol / 0.25 mol
Mole fraction of O2 = 0.2
Partial pressure of O2 = Mole fraction of O2 x Total pressure
Partial pressure of O2 = 0.2 x 0.768 atm
Partial pressure of O2 = 0.1536 atm
Therefore, the partial pressure of O2 in atm is 0.1536 atm, which is between the given range of 0.06-0.07 atm.
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Community Answer
A 10 L flask containing 10.8 g of N2O5 is heated to 373 K, which leads...
V=10L
T=373K
R=0.0821
P=0.5 atm
PV=nRT
0.5×10=n×0.0821×373
n=0.163274
here n is the total no. of moles of the solution at equilibrium or time t.
Now,
given mass of N2O5=10.8
molar mass of N2O5=108
Moles of N205=0.1
2N205 -. 4NO2. + O2
At t=0. 0.1. 0. 0.
At t=t. 0.1-2x. 4x x
A/Q=
0.1-2x+4x+x= Total moles of the given solution (0.163274)
0.1+3x= 0.163274
3x=0.063274
x= 0.02109
Now mole fraction of O2= 0.02109÷ 0.162374= 0.129172
Partial pressure of O2 =Mole fraction of O2× Total pressure
=0.129712×0.5
=0.06458 atm
So. partial pressure of O2 is 0.06458 atm
T=373K
R=0.0821
P=0.5 atm
PV=nRT
0.5×10=n×0.0821×373
n=0.163274
here n is the total no. of moles of the solution at equilibrium or time t.
Now,
given mass of N2O5=10.8
molar mass of N2O5=108
Moles of N205=0.1
2N205 -. 4NO2. + O2
At t=0. 0.1. 0. 0.
At t=t. 0.1-2x. 4x x
A/Q=
0.1-2x+4x+x= Total moles of the given solution (0.163274)
0.1+3x= 0.163274
3x=0.063274
x= 0.02109
Now mole fraction of O2= 0.02109÷ 0.162374= 0.129172
Partial pressure of O2 =Mole fraction of O2× Total pressure
=0.129712×0.5
=0.06458 atm
So. partial pressure of O2 is 0.06458 atm
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A 10 L flask containing 10.8 g of N2O5 is heated to 373 K, which leads to its decomposition according to the equation 2 N2O5(g) → 4 NO2(g) + O2(g). If the final pressure in the flask is 0.5 atm, then the partial pressure of O2(g) in atm is __.[Given R = 0.0821 L atm K–1 mol–1]Correct answer is between '0.06,0.07'. Can you explain this answer?
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A 10 L flask containing 10.8 g of N2O5 is heated to 373 K, which leads to its decomposition according to the equation 2 N2O5(g) → 4 NO2(g) + O2(g). If the final pressure in the flask is 0.5 atm, then the partial pressure of O2(g) in atm is __.[Given R = 0.0821 L atm K–1 mol–1]Correct answer is between '0.06,0.07'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about A 10 L flask containing 10.8 g of N2O5 is heated to 373 K, which leads to its decomposition according to the equation 2 N2O5(g) → 4 NO2(g) + O2(g). If the final pressure in the flask is 0.5 atm, then the partial pressure of O2(g) in atm is __.[Given R = 0.0821 L atm K–1 mol–1]Correct answer is between '0.06,0.07'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 10 L flask containing 10.8 g of N2O5 is heated to 373 K, which leads to its decomposition according to the equation 2 N2O5(g) → 4 NO2(g) + O2(g). If the final pressure in the flask is 0.5 atm, then the partial pressure of O2(g) in atm is __.[Given R = 0.0821 L atm K–1 mol–1]Correct answer is between '0.06,0.07'. Can you explain this answer?.
A 10 L flask containing 10.8 g of N2O5 is heated to 373 K, which leads to its decomposition according to the equation 2 N2O5(g) → 4 NO2(g) + O2(g). If the final pressure in the flask is 0.5 atm, then the partial pressure of O2(g) in atm is __.[Given R = 0.0821 L atm K–1 mol–1]Correct answer is between '0.06,0.07'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about A 10 L flask containing 10.8 g of N2O5 is heated to 373 K, which leads to its decomposition according to the equation 2 N2O5(g) → 4 NO2(g) + O2(g). If the final pressure in the flask is 0.5 atm, then the partial pressure of O2(g) in atm is __.[Given R = 0.0821 L atm K–1 mol–1]Correct answer is between '0.06,0.07'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 10 L flask containing 10.8 g of N2O5 is heated to 373 K, which leads to its decomposition according to the equation 2 N2O5(g) → 4 NO2(g) + O2(g). If the final pressure in the flask is 0.5 atm, then the partial pressure of O2(g) in atm is __.[Given R = 0.0821 L atm K–1 mol–1]Correct answer is between '0.06,0.07'. Can you explain this answer?.
Solutions for A 10 L flask containing 10.8 g of N2O5 is heated to 373 K, which leads to its decomposition according to the equation 2 N2O5(g) → 4 NO2(g) + O2(g). If the final pressure in the flask is 0.5 atm, then the partial pressure of O2(g) in atm is __.[Given R = 0.0821 L atm K–1 mol–1]Correct answer is between '0.06,0.07'. Can you explain this answer? in English & in Hindi are available as part of our courses for Chemistry.
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A 10 L flask containing 10.8 g of N2O5 is heated to 373 K, which leads to its decomposition according to the equation 2 N2O5(g) → 4 NO2(g) + O2(g). If the final pressure in the flask is 0.5 atm, then the partial pressure of O2(g) in atm is __.[Given R = 0.0821 L atm K–1 mol–1]Correct answer is between '0.06,0.07'. Can you explain this answer?, a detailed solution for A 10 L flask containing 10.8 g of N2O5 is heated to 373 K, which leads to its decomposition according to the equation 2 N2O5(g) → 4 NO2(g) + O2(g). If the final pressure in the flask is 0.5 atm, then the partial pressure of O2(g) in atm is __.[Given R = 0.0821 L atm K–1 mol–1]Correct answer is between '0.06,0.07'. Can you explain this answer? has been provided alongside types of A 10 L flask containing 10.8 g of N2O5 is heated to 373 K, which leads to its decomposition according to the equation 2 N2O5(g) → 4 NO2(g) + O2(g). If the final pressure in the flask is 0.5 atm, then the partial pressure of O2(g) in atm is __.[Given R = 0.0821 L atm K–1 mol–1]Correct answer is between '0.06,0.07'. Can you explain this answer? theory, EduRev gives you an
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