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A mixture of C3H8 and oxygen in 1L closed vessel has an internal pressure of 4 atm at 100ºC. When the mixture is ignited, the reaction produces CO2(g) and H2O(g) until all oxygen is consumed.
After the reaction, pressure of the vessel is 4.2 atm at the same temperature. Calculate the weight of oxygen present before the reaction. [Gas constant, R = 0.082 L atm mol–1 K–1].
After the reaction, pressure of the vessel is 4.2 atm at the same temperature. Calculate the weight of oxygen present before the reaction. [Gas constant, R = 0.082 L atm mol–1 K–1].
Correct answer is 'C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)'. Can you explain this answer?
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Verified Answer
A mixture of C3H8 and oxygen in 1L closed vessel has an internal press...
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
Initial number of moles, n =
= 0.1306Final number of moles, nf
= 
x = 5 × 0.0064 ⇒ x = 0.032
weight of O2 = x × 32 = 0.032 × 32 = 1.024 gm
weight of O2 = x × 32 = 0.032 × 32 = 1.024 gm
Most Upvoted Answer
A mixture of C3H8 and oxygen in 1L closed vessel has an internal press...
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
Initial number of moles, n =
= 0.1306Final number of moles, nf
= 
x = 5 × 0.0064 ⇒ x = 0.032
weight of O2 = x × 32 = 0.032 × 32 = 1.024 gm
weight of O2 = x × 32 = 0.032 × 32 = 1.024 gm
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A mixture of C3H8 and oxygen in 1L closed vessel has an internal press...
Given information:
- Mixture of C3H8 and oxygen in a 1L closed vessel
- Internal pressure of the mixture is 4 atm at 100°C
- Reaction produces CO2(g) and H2O(g) until all oxygen is consumed
- Pressure of the vessel after the reaction is 4.2 atm at the same temperature
- Gas constant, R = 0.082 L atm mol^-1 K^-1
Equation for the reaction:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
Solution:
To find the weight of oxygen present before the reaction, we need to calculate the number of moles of oxygen before and after the reaction and then convert it to grams.
Step 1: Calculate the number of moles of oxygen before the reaction
We can use the ideal gas law to calculate the number of moles of oxygen before the reaction using the given information.
PV = nRT
Where:
P = pressure = 4 atm
V = volume = 1 L
n = number of moles of oxygen
R = gas constant = 0.082 L atm mol^-1 K^-1
T = temperature = 100°C = 373 K
Rearranging the equation, we can solve for n:
n = PV / RT
Substituting the values:
n = (4 atm)(1 L) / (0.082 L atm mol^-1 K^-1)(373 K)
Calculating, we get:
n = 0.129 moles
Step 2: Calculate the weight of oxygen before the reaction
To calculate the weight of oxygen, we need to use the molar mass of oxygen, which is 32 g/mol.
Weight = number of moles × molar mass
Weight = 0.129 moles × 32 g/mol
Calculating, we get:
Weight = 4.128 g
Therefore, the weight of oxygen present before the reaction is 4.128 grams.
- Mixture of C3H8 and oxygen in a 1L closed vessel
- Internal pressure of the mixture is 4 atm at 100°C
- Reaction produces CO2(g) and H2O(g) until all oxygen is consumed
- Pressure of the vessel after the reaction is 4.2 atm at the same temperature
- Gas constant, R = 0.082 L atm mol^-1 K^-1
Equation for the reaction:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
Solution:
To find the weight of oxygen present before the reaction, we need to calculate the number of moles of oxygen before and after the reaction and then convert it to grams.
Step 1: Calculate the number of moles of oxygen before the reaction
We can use the ideal gas law to calculate the number of moles of oxygen before the reaction using the given information.
PV = nRT
Where:
P = pressure = 4 atm
V = volume = 1 L
n = number of moles of oxygen
R = gas constant = 0.082 L atm mol^-1 K^-1
T = temperature = 100°C = 373 K
Rearranging the equation, we can solve for n:
n = PV / RT
Substituting the values:
n = (4 atm)(1 L) / (0.082 L atm mol^-1 K^-1)(373 K)
Calculating, we get:
n = 0.129 moles
Step 2: Calculate the weight of oxygen before the reaction
To calculate the weight of oxygen, we need to use the molar mass of oxygen, which is 32 g/mol.
Weight = number of moles × molar mass
Weight = 0.129 moles × 32 g/mol
Calculating, we get:
Weight = 4.128 g
Therefore, the weight of oxygen present before the reaction is 4.128 grams.
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A mixture of C3H8 and oxygen in 1L closed vessel has an internal pressure of 4 atm at 100C. When the mixture is ignited, the reaction produces CO2(g) and H2O(g) until all oxygen is consumed.After the reaction, pressure of the vessel is 4.2 atm at the same temperature. Calculate the weight of oxygen present before the reaction. [Gas constant, R = 0.082 L atm mol1 K1].Correct answer is 'C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)'. Can you explain this answer?
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A mixture of C3H8 and oxygen in 1L closed vessel has an internal pressure of 4 atm at 100C. When the mixture is ignited, the reaction produces CO2(g) and H2O(g) until all oxygen is consumed.After the reaction, pressure of the vessel is 4.2 atm at the same temperature. Calculate the weight of oxygen present before the reaction. [Gas constant, R = 0.082 L atm mol1 K1].Correct answer is 'C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about A mixture of C3H8 and oxygen in 1L closed vessel has an internal pressure of 4 atm at 100C. When the mixture is ignited, the reaction produces CO2(g) and H2O(g) until all oxygen is consumed.After the reaction, pressure of the vessel is 4.2 atm at the same temperature. Calculate the weight of oxygen present before the reaction. [Gas constant, R = 0.082 L atm mol1 K1].Correct answer is 'C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A mixture of C3H8 and oxygen in 1L closed vessel has an internal pressure of 4 atm at 100C. When the mixture is ignited, the reaction produces CO2(g) and H2O(g) until all oxygen is consumed.After the reaction, pressure of the vessel is 4.2 atm at the same temperature. Calculate the weight of oxygen present before the reaction. [Gas constant, R = 0.082 L atm mol1 K1].Correct answer is 'C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)'. Can you explain this answer?.
A mixture of C3H8 and oxygen in 1L closed vessel has an internal pressure of 4 atm at 100C. When the mixture is ignited, the reaction produces CO2(g) and H2O(g) until all oxygen is consumed.After the reaction, pressure of the vessel is 4.2 atm at the same temperature. Calculate the weight of oxygen present before the reaction. [Gas constant, R = 0.082 L atm mol1 K1].Correct answer is 'C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about A mixture of C3H8 and oxygen in 1L closed vessel has an internal pressure of 4 atm at 100C. When the mixture is ignited, the reaction produces CO2(g) and H2O(g) until all oxygen is consumed.After the reaction, pressure of the vessel is 4.2 atm at the same temperature. Calculate the weight of oxygen present before the reaction. [Gas constant, R = 0.082 L atm mol1 K1].Correct answer is 'C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A mixture of C3H8 and oxygen in 1L closed vessel has an internal pressure of 4 atm at 100C. When the mixture is ignited, the reaction produces CO2(g) and H2O(g) until all oxygen is consumed.After the reaction, pressure of the vessel is 4.2 atm at the same temperature. Calculate the weight of oxygen present before the reaction. [Gas constant, R = 0.082 L atm mol1 K1].Correct answer is 'C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)'. Can you explain this answer?.
Solutions for A mixture of C3H8 and oxygen in 1L closed vessel has an internal pressure of 4 atm at 100C. When the mixture is ignited, the reaction produces CO2(g) and H2O(g) until all oxygen is consumed.After the reaction, pressure of the vessel is 4.2 atm at the same temperature. Calculate the weight of oxygen present before the reaction. [Gas constant, R = 0.082 L atm mol1 K1].Correct answer is 'C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)'. Can you explain this answer? in English & in Hindi are available as part of our courses for IIT JAM.
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