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1.00 L sample of a mixture of methane gas and oxygen gas measured at 25°C and 740 torr was allowed to react at constant pressure in a calorimeter. The calorimeter together with its contents had a heat capacity of 1000 cal/K. The complete combustion of methane to CO2 and water caused a rise in temperature of 0.42 K. Heat of the following reaction is ΔH = - 210.8 kcal.
CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (I) Mole percentage of methane in the original mixture is
Correct answer is '5'. Can you explain this answer?
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1.00 L sample of a mixture of methane gas and oxygen gas measured at ...
Mole Percentage of Methane in the Mixture
Given:
- Volume of the mixture = 1.00 L
- Temperature = 25°C = 25 + 273 = 298 K
- Pressure = 740 torr
- Heat capacity of the calorimeter and its contents = 1000 cal/K
- Rise in temperature = 0.42 K
- Heat of the reaction (ΔH) = -210.8 kcal
To find the mole percentage of methane in the original mixture, we need to calculate the number of moles of methane and the number of moles of oxygen in the mixture.
Step 1: Convert temperature and pressure to the appropriate units
- Convert temperature from °C to Kelvin: 25 + 273 = 298 K
- Convert pressure from torr to atm: 740 torr ÷ 760 = 0.974 atm
Step 2: Use the ideal gas law to calculate the number of moles of each gas
- Ideal gas law equation: PV = nRT
- Rearrange the equation to solve for n (number of moles): n = PV / RT
For methane:
- P = 0.974 atm
- V = 1.00 L
- R (universal gas constant) = 0.0821 L·atm/(K·mol)
- T = 298 K
nCH4 = (0.974 atm * 1.00 L) / (0.0821 L·atm/(K·mol) * 298 K) ≈ 0.0413 mol
For oxygen:
- Since the reaction is 2 moles of oxygen per mole of methane, the number of moles of oxygen is twice the number of moles of methane.
nO2 = 2 * nCH4 ≈ 2 * 0.0413 mol ≈ 0.0826 mol
Step 3: Calculate the heat released in the reaction
- The heat released in the reaction can be calculated using the equation: q = ΔH * n
q = -210.8 kcal * 0.0413 mol ≈ -8.71 kcal
Step 4: Calculate the heat absorbed by the calorimeter and its contents
- The heat absorbed by the calorimeter and its contents can be calculated using the equation: q = C * ΔT
- C = heat capacity of the calorimeter and its contents = 1000 cal/K
- ΔT = rise in temperature = 0.42 K
q = 1000 cal/K * 0.42 K = 420 cal
Step 5: Calculate the heat absorbed by the surroundings (water)
- The heat absorbed by the surroundings can be calculated using the equation: q = -q
- The heat absorbed by the surroundings is equal in magnitude but opposite in sign to the heat released in the reaction.
q = -420 cal
Step 6: Calculate the heat absorbed by the water
- The heat absorbed by the water can be calculated using the equation: q = m * C * ΔT
- m = mass of water
- C = specific heat capacity of water = 1 cal/g·K
- ΔT = rise in temperature = 0.42 K
q = m * 1 cal/g·K * 0.42 K
Since the mass
Given:
- Volume of the mixture = 1.00 L
- Temperature = 25°C = 25 + 273 = 298 K
- Pressure = 740 torr
- Heat capacity of the calorimeter and its contents = 1000 cal/K
- Rise in temperature = 0.42 K
- Heat of the reaction (ΔH) = -210.8 kcal
To find the mole percentage of methane in the original mixture, we need to calculate the number of moles of methane and the number of moles of oxygen in the mixture.
Step 1: Convert temperature and pressure to the appropriate units
- Convert temperature from °C to Kelvin: 25 + 273 = 298 K
- Convert pressure from torr to atm: 740 torr ÷ 760 = 0.974 atm
Step 2: Use the ideal gas law to calculate the number of moles of each gas
- Ideal gas law equation: PV = nRT
- Rearrange the equation to solve for n (number of moles): n = PV / RT
For methane:
- P = 0.974 atm
- V = 1.00 L
- R (universal gas constant) = 0.0821 L·atm/(K·mol)
- T = 298 K
nCH4 = (0.974 atm * 1.00 L) / (0.0821 L·atm/(K·mol) * 298 K) ≈ 0.0413 mol
For oxygen:
- Since the reaction is 2 moles of oxygen per mole of methane, the number of moles of oxygen is twice the number of moles of methane.
nO2 = 2 * nCH4 ≈ 2 * 0.0413 mol ≈ 0.0826 mol
Step 3: Calculate the heat released in the reaction
- The heat released in the reaction can be calculated using the equation: q = ΔH * n
q = -210.8 kcal * 0.0413 mol ≈ -8.71 kcal
Step 4: Calculate the heat absorbed by the calorimeter and its contents
- The heat absorbed by the calorimeter and its contents can be calculated using the equation: q = C * ΔT
- C = heat capacity of the calorimeter and its contents = 1000 cal/K
- ΔT = rise in temperature = 0.42 K
q = 1000 cal/K * 0.42 K = 420 cal
Step 5: Calculate the heat absorbed by the surroundings (water)
- The heat absorbed by the surroundings can be calculated using the equation: q = -q
- The heat absorbed by the surroundings is equal in magnitude but opposite in sign to the heat released in the reaction.
q = -420 cal
Step 6: Calculate the heat absorbed by the water
- The heat absorbed by the water can be calculated using the equation: q = m * C * ΔT
- m = mass of water
- C = specific heat capacity of water = 1 cal/g·K
- ΔT = rise in temperature = 0.42 K
q = m * 1 cal/g·K * 0.42 K
Since the mass
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Community Answer
1.00 L sample of a mixture of methane gas and oxygen gas measured at ...
Heat generated = (0.42 K × 1000 cal/K) = 420 cal
Number of moles of Methane gas in the mixture = 420 cal x 
= 1.99 10-3 mol
Ntotal = 
= 0.0398 mol
Mole percentage of methane = 
= 5
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1.00 L sample of a mixture of methane gas and oxygen gas measured at 25°C and 740 torr was allowed to react at constant pressure in a calorimeter. The calorimeter together with its contents had a heat capacity of 1000 cal/K. The complete combustion of methane to CO2 and water caused a rise in temperature of 0.42 K. Heat of the following reaction is ΔH = - 210.8 kcal.CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (I) Mole percentage of methane in the original mixture isCorrect answer is '5'. Can you explain this answer?
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1.00 L sample of a mixture of methane gas and oxygen gas measured at 25°C and 740 torr was allowed to react at constant pressure in a calorimeter. The calorimeter together with its contents had a heat capacity of 1000 cal/K. The complete combustion of methane to CO2 and water caused a rise in temperature of 0.42 K. Heat of the following reaction is ΔH = - 210.8 kcal.CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (I) Mole percentage of methane in the original mixture isCorrect answer is '5'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 1.00 L sample of a mixture of methane gas and oxygen gas measured at 25°C and 740 torr was allowed to react at constant pressure in a calorimeter. The calorimeter together with its contents had a heat capacity of 1000 cal/K. The complete combustion of methane to CO2 and water caused a rise in temperature of 0.42 K. Heat of the following reaction is ΔH = - 210.8 kcal.CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (I) Mole percentage of methane in the original mixture isCorrect answer is '5'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1.00 L sample of a mixture of methane gas and oxygen gas measured at 25°C and 740 torr was allowed to react at constant pressure in a calorimeter. The calorimeter together with its contents had a heat capacity of 1000 cal/K. The complete combustion of methane to CO2 and water caused a rise in temperature of 0.42 K. Heat of the following reaction is ΔH = - 210.8 kcal.CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (I) Mole percentage of methane in the original mixture isCorrect answer is '5'. Can you explain this answer?.
1.00 L sample of a mixture of methane gas and oxygen gas measured at 25°C and 740 torr was allowed to react at constant pressure in a calorimeter. The calorimeter together with its contents had a heat capacity of 1000 cal/K. The complete combustion of methane to CO2 and water caused a rise in temperature of 0.42 K. Heat of the following reaction is ΔH = - 210.8 kcal.CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (I) Mole percentage of methane in the original mixture isCorrect answer is '5'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 1.00 L sample of a mixture of methane gas and oxygen gas measured at 25°C and 740 torr was allowed to react at constant pressure in a calorimeter. The calorimeter together with its contents had a heat capacity of 1000 cal/K. The complete combustion of methane to CO2 and water caused a rise in temperature of 0.42 K. Heat of the following reaction is ΔH = - 210.8 kcal.CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (I) Mole percentage of methane in the original mixture isCorrect answer is '5'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1.00 L sample of a mixture of methane gas and oxygen gas measured at 25°C and 740 torr was allowed to react at constant pressure in a calorimeter. The calorimeter together with its contents had a heat capacity of 1000 cal/K. The complete combustion of methane to CO2 and water caused a rise in temperature of 0.42 K. Heat of the following reaction is ΔH = - 210.8 kcal.CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (I) Mole percentage of methane in the original mixture isCorrect answer is '5'. Can you explain this answer?.
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1.00 L sample of a mixture of methane gas and oxygen gas measured at 25°C and 740 torr was allowed to react at constant pressure in a calorimeter. The calorimeter together with its contents had a heat capacity of 1000 cal/K. The complete combustion of methane to CO2 and water caused a rise in temperature of 0.42 K. Heat of the following reaction is ΔH = - 210.8 kcal.CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (I) Mole percentage of methane in the original mixture isCorrect answer is '5'. Can you explain this answer?, a detailed solution for 1.00 L sample of a mixture of methane gas and oxygen gas measured at 25°C and 740 torr was allowed to react at constant pressure in a calorimeter. The calorimeter together with its contents had a heat capacity of 1000 cal/K. The complete combustion of methane to CO2 and water caused a rise in temperature of 0.42 K. Heat of the following reaction is ΔH = - 210.8 kcal.CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (I) Mole percentage of methane in the original mixture isCorrect answer is '5'. Can you explain this answer? has been provided alongside types of 1.00 L sample of a mixture of methane gas and oxygen gas measured at 25°C and 740 torr was allowed to react at constant pressure in a calorimeter. The calorimeter together with its contents had a heat capacity of 1000 cal/K. The complete combustion of methane to CO2 and water caused a rise in temperature of 0.42 K. Heat of the following reaction is ΔH = - 210.8 kcal.CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (I) Mole percentage of methane in the original mixture isCorrect answer is '5'. Can you explain this answer? theory, EduRev gives you an
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