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A mixture of C3H8 and oxygen in 1 L closed vessel has an internal pressure of 4 atm at 100°C. When the mixture is ignited, the reaction produces CO2(g) and H2O(g) until all the 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] (upto 2 decimal place):
Correct answer is between '0.90,1.00'. Can you explain this answer?
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A mixture of C3H8 and oxygen in 1 L closed vessel has an internal pres...
°C. If all the propane is burned completely to form CO2 and H2O, what is the final pressure in the vessel?
The balanced chemical equation for the combustion of propane is:
C3H8 + 5O2 → 3CO2 + 4H2O
From the equation, we can see that every mole of propane (C3H8) requires 5 moles of oxygen (O2) to produce 3 moles of carbon dioxide (CO2) and 4 moles of water (H2O). Therefore, we need to determine the number of moles of propane and oxygen in the 1 L closed vessel.
Using the ideal gas law:
PV = nRT
Where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.
We can rearrange the formula to solve for n:
n = PV/RT
n(propane) = (4 atm)(1 L)/(0.0821 L·atm/mol·K)(373 K) = 0.131 mol
n(oxygen) = (4 atm)(1 L)/(0.0821 L·atm/mol·K)(373 K) = 0.656 mol
Therefore, we have 0.131 mol of propane and 0.656 mol of oxygen in the vessel. Since all the propane will react with the oxygen to produce CO2 and H2O, we can use the stoichiometry of the balanced equation to determine the number of moles of products.
From the equation, every 1 mole of propane produces 3 moles of CO2 and 4 moles of H2O. Therefore, the 0.131 mol of propane will produce:
3 mol CO2/mol propane × 0.131 mol propane = 0.393 mol CO2
4 mol H2O/mol propane × 0.131 mol propane = 0.524 mol H2O
Since the volume and temperature of the vessel remain constant, we can use the ideal gas law again to determine the final pressure of the system. We can assume that the CO2 and H2O gases behave ideally.
n(total) = n(propane) + n(oxygen) + n(CO2) + n(H2O)
n(total) = 0.131 mol + 0.656 mol + 0.393 mol + 0.524 mol
n(total) = 1.704 mol
P(total) = n(total)RT/V
P(total) = (1.704 mol)(0.0821 L·atm/mol·K)(373 K)/(1 L)
P(total) = 5.59 atm
Therefore, the final pressure in the vessel after the combustion of propane is 5.59 atm.
The balanced chemical equation for the combustion of propane is:
C3H8 + 5O2 → 3CO2 + 4H2O
From the equation, we can see that every mole of propane (C3H8) requires 5 moles of oxygen (O2) to produce 3 moles of carbon dioxide (CO2) and 4 moles of water (H2O). Therefore, we need to determine the number of moles of propane and oxygen in the 1 L closed vessel.
Using the ideal gas law:
PV = nRT
Where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.
We can rearrange the formula to solve for n:
n = PV/RT
n(propane) = (4 atm)(1 L)/(0.0821 L·atm/mol·K)(373 K) = 0.131 mol
n(oxygen) = (4 atm)(1 L)/(0.0821 L·atm/mol·K)(373 K) = 0.656 mol
Therefore, we have 0.131 mol of propane and 0.656 mol of oxygen in the vessel. Since all the propane will react with the oxygen to produce CO2 and H2O, we can use the stoichiometry of the balanced equation to determine the number of moles of products.
From the equation, every 1 mole of propane produces 3 moles of CO2 and 4 moles of H2O. Therefore, the 0.131 mol of propane will produce:
3 mol CO2/mol propane × 0.131 mol propane = 0.393 mol CO2
4 mol H2O/mol propane × 0.131 mol propane = 0.524 mol H2O
Since the volume and temperature of the vessel remain constant, we can use the ideal gas law again to determine the final pressure of the system. We can assume that the CO2 and H2O gases behave ideally.
n(total) = n(propane) + n(oxygen) + n(CO2) + n(H2O)
n(total) = 0.131 mol + 0.656 mol + 0.393 mol + 0.524 mol
n(total) = 1.704 mol
P(total) = n(total)RT/V
P(total) = (1.704 mol)(0.0821 L·atm/mol·K)(373 K)/(1 L)
P(total) = 5.59 atm
Therefore, the final pressure in the vessel after the combustion of propane is 5.59 atm.
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A mixture of C3H8 and oxygen in 1 L closed vessel has an internal pressure of 4 atm at 100°C. When the mixture is ignited, the reaction produces CO2(g) and H2O(g) until all the 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](upto 2 decimal place):Correct answer is between '0.90,1.00'. Can you explain this answer?
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A mixture of C3H8 and oxygen in 1 L closed vessel has an internal pressure of 4 atm at 100°C. When the mixture is ignited, the reaction produces CO2(g) and H2O(g) until all the 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](upto 2 decimal place):Correct answer is between '0.90,1.00'. 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 1 L closed vessel has an internal pressure of 4 atm at 100°C. When the mixture is ignited, the reaction produces CO2(g) and H2O(g) until all the 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](upto 2 decimal place):Correct answer is between '0.90,1.00'. 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 1 L closed vessel has an internal pressure of 4 atm at 100°C. When the mixture is ignited, the reaction produces CO2(g) and H2O(g) until all the 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](upto 2 decimal place):Correct answer is between '0.90,1.00'. Can you explain this answer?.
A mixture of C3H8 and oxygen in 1 L closed vessel has an internal pressure of 4 atm at 100°C. When the mixture is ignited, the reaction produces CO2(g) and H2O(g) until all the 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](upto 2 decimal place):Correct answer is between '0.90,1.00'. 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 1 L closed vessel has an internal pressure of 4 atm at 100°C. When the mixture is ignited, the reaction produces CO2(g) and H2O(g) until all the 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](upto 2 decimal place):Correct answer is between '0.90,1.00'. 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 1 L closed vessel has an internal pressure of 4 atm at 100°C. When the mixture is ignited, the reaction produces CO2(g) and H2O(g) until all the 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](upto 2 decimal place):Correct answer is between '0.90,1.00'. Can you explain this answer?.
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