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The heat evolved in the combustion of methane is given by the following equations:
CH4 (g) + 2O2 (g) -> CO2 (g) + H2O (l) ΔH = -890.3 KJ
How many grams of methane would be required to produce 444.15 KJ of heat of combustion:
- a)4 g
- b)8 g
- c)12 g
- d)16 g
Correct answer is option 'B'. Can you explain this answer?
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The heat evolved in the combustion of methane is given by the followin...
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The heat evolved in the combustion of methane is given by the followin...
→ CO2 (g) 2H2O (l)
The balanced equation for the combustion of methane is:
CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l)
In this reaction, one mole of methane (CH4) reacts with two moles of oxygen (O2) to produce one mole of carbon dioxide (CO2) and two moles of water (H2O).
The heat evolved in the combustion of methane can be calculated using the enthalpy change of formation (ΔHf) values for the reactants and products. The ΔHf values represent the heat change that occurs when one mole of a substance is formed from its elements in their standard states.
The standard enthalpy change of formation for methane (ΔHf(CH4)) is -74.8 kJ/mol.
The standard enthalpy change of formation for carbon dioxide (ΔHf(CO2)) is -393.5 kJ/mol.
The standard enthalpy change of formation for water (ΔHf(H2O)) is -285.8 kJ/mol.
To calculate the heat evolved in the combustion of methane, we can use the following equation:
ΔHrxn = ΣΔHf(products) - ΣΔHf(reactants)
ΔHrxn = [ΔHf(CO2)] + 2[ΔHf(H2O)] - [ΔHf(CH4)]
ΔHrxn = (-393.5 kJ/mol) + 2(-285.8 kJ/mol) - (-74.8 kJ/mol)
ΔHrxn = -393.5 kJ/mol - 571.6 kJ/mol + 74.8 kJ/mol
ΔHrxn = -890.3 kJ/mol
Therefore, the heat evolved in the combustion of one mole of methane is -890.3 kJ.
The balanced equation for the combustion of methane is:
CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l)
In this reaction, one mole of methane (CH4) reacts with two moles of oxygen (O2) to produce one mole of carbon dioxide (CO2) and two moles of water (H2O).
The heat evolved in the combustion of methane can be calculated using the enthalpy change of formation (ΔHf) values for the reactants and products. The ΔHf values represent the heat change that occurs when one mole of a substance is formed from its elements in their standard states.
The standard enthalpy change of formation for methane (ΔHf(CH4)) is -74.8 kJ/mol.
The standard enthalpy change of formation for carbon dioxide (ΔHf(CO2)) is -393.5 kJ/mol.
The standard enthalpy change of formation for water (ΔHf(H2O)) is -285.8 kJ/mol.
To calculate the heat evolved in the combustion of methane, we can use the following equation:
ΔHrxn = ΣΔHf(products) - ΣΔHf(reactants)
ΔHrxn = [ΔHf(CO2)] + 2[ΔHf(H2O)] - [ΔHf(CH4)]
ΔHrxn = (-393.5 kJ/mol) + 2(-285.8 kJ/mol) - (-74.8 kJ/mol)
ΔHrxn = -393.5 kJ/mol - 571.6 kJ/mol + 74.8 kJ/mol
ΔHrxn = -890.3 kJ/mol
Therefore, the heat evolved in the combustion of one mole of methane is -890.3 kJ.
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The heat evolved in the combustion of methane is given by the following equations:CH4 (g) + 2O2 (g) -> CO2 (g) + H2O (l) ΔH = -890.3 KJHow many grams of methane would be required to produce 444.15 KJ of heat of combustion:a)4 g b)8 g c)12 g d)16 gCorrect answer is option 'B'. Can you explain this answer?
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The heat evolved in the combustion of methane is given by the following equations:CH4 (g) + 2O2 (g) -> CO2 (g) + H2O (l) ΔH = -890.3 KJHow many grams of methane would be required to produce 444.15 KJ of heat of combustion:a)4 g b)8 g c)12 g d)16 gCorrect answer is option 'B'. 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 The heat evolved in the combustion of methane is given by the following equations:CH4 (g) + 2O2 (g) -> CO2 (g) + H2O (l) ΔH = -890.3 KJHow many grams of methane would be required to produce 444.15 KJ of heat of combustion:a)4 g b)8 g c)12 g d)16 gCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The heat evolved in the combustion of methane is given by the following equations:CH4 (g) + 2O2 (g) -> CO2 (g) + H2O (l) ΔH = -890.3 KJHow many grams of methane would be required to produce 444.15 KJ of heat of combustion:a)4 g b)8 g c)12 g d)16 gCorrect answer is option 'B'. Can you explain this answer?.
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