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If heat of combustion of ethylene is 1411kj when a certain amount of ethylene was burnt 6226kj heat was evolved.then the volume of o2(at NTP)that entered in to the rxn is? 1》296.5ml 2》296.5 litre 3》6226×22.4 litre4》22.4 litre.plz explain with solition?
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If heat of combustion of ethylene is 1411kj when a certain amount of e...
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If heat of combustion of ethylene is 1411kj when a certain amount of e...
Solution:
Given:
Heat of combustion of ethylene = 1411 kJ
Heat evolved = 6226 kJ
To find:
Volume of O2 at NTP that entered the reaction
We can solve this problem using the concept of stoichiometry. The balanced equation for the combustion of ethylene (C2H4) is:
C2H4 + O2 → CO2 + H2O
From the balanced equation, we can see that the stoichiometric ratio between ethylene and oxygen is 1:3. This means that for every 1 mole of ethylene burned, 3 moles of oxygen are required.
Step 1: Calculate the moles of ethylene burned.
To calculate the moles of ethylene, we use the formula:
moles = heat evolved / heat of combustion
moles of ethylene = 6226 kJ / 1411 kJ/mol = 4.41 mol
Step 2: Calculate the moles of oxygen required.
Since the stoichiometric ratio of ethylene to oxygen is 1:3, the moles of oxygen required can be calculated as follows:
moles of oxygen = 3 * moles of ethylene
moles of oxygen = 3 * 4.41 mol = 13.23 mol
Step 3: Calculate the volume of oxygen at NTP.
At NTP (normal temperature and pressure), 1 mole of any gas occupies 22.4 liters of volume. Therefore, the volume of oxygen can be calculated as:
volume of oxygen = moles of oxygen * 22.4 L/mol
volume of oxygen = 13.23 mol * 22.4 L/mol = 296.5 liters
Answer: The volume of oxygen (at NTP) that entered the reaction is 296.5 liters.
Summary:
To find the volume of oxygen at NTP that entered the reaction, we used the stoichiometry of the balanced equation. The moles of ethylene burned were determined using the heat evolved and the heat of combustion. From there, we calculated the moles of oxygen required and converted it to volume using the molar volume at NTP. The final answer obtained was 296.5 liters.
Given:
Heat of combustion of ethylene = 1411 kJ
Heat evolved = 6226 kJ
To find:
Volume of O2 at NTP that entered the reaction
We can solve this problem using the concept of stoichiometry. The balanced equation for the combustion of ethylene (C2H4) is:
C2H4 + O2 → CO2 + H2O
From the balanced equation, we can see that the stoichiometric ratio between ethylene and oxygen is 1:3. This means that for every 1 mole of ethylene burned, 3 moles of oxygen are required.
Step 1: Calculate the moles of ethylene burned.
To calculate the moles of ethylene, we use the formula:
moles = heat evolved / heat of combustion
moles of ethylene = 6226 kJ / 1411 kJ/mol = 4.41 mol
Step 2: Calculate the moles of oxygen required.
Since the stoichiometric ratio of ethylene to oxygen is 1:3, the moles of oxygen required can be calculated as follows:
moles of oxygen = 3 * moles of ethylene
moles of oxygen = 3 * 4.41 mol = 13.23 mol
Step 3: Calculate the volume of oxygen at NTP.
At NTP (normal temperature and pressure), 1 mole of any gas occupies 22.4 liters of volume. Therefore, the volume of oxygen can be calculated as:
volume of oxygen = moles of oxygen * 22.4 L/mol
volume of oxygen = 13.23 mol * 22.4 L/mol = 296.5 liters
Answer: The volume of oxygen (at NTP) that entered the reaction is 296.5 liters.
Summary:
To find the volume of oxygen at NTP that entered the reaction, we used the stoichiometry of the balanced equation. The moles of ethylene burned were determined using the heat evolved and the heat of combustion. From there, we calculated the moles of oxygen required and converted it to volume using the molar volume at NTP. The final answer obtained was 296.5 liters.
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If heat of combustion of ethylene is 1411kj when a certain amount of ethylene was burnt 6226kj heat was evolved.then the volume of o2(at NTP)that entered in to the rxn is? 1》296.5ml 2》296.5 litre 3》6226×22.4 litre4》22.4 litre.plz explain with solition?
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If heat of combustion of ethylene is 1411kj when a certain amount of ethylene was burnt 6226kj heat was evolved.then the volume of o2(at NTP)that entered in to the rxn is? 1》296.5ml 2》296.5 litre 3》6226×22.4 litre4》22.4 litre.plz explain with solition? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about If heat of combustion of ethylene is 1411kj when a certain amount of ethylene was burnt 6226kj heat was evolved.then the volume of o2(at NTP)that entered in to the rxn is? 1》296.5ml 2》296.5 litre 3》6226×22.4 litre4》22.4 litre.plz explain with solition? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If heat of combustion of ethylene is 1411kj when a certain amount of ethylene was burnt 6226kj heat was evolved.then the volume of o2(at NTP)that entered in to the rxn is? 1》296.5ml 2》296.5 litre 3》6226×22.4 litre4》22.4 litre.plz explain with solition?.
If heat of combustion of ethylene is 1411kj when a certain amount of ethylene was burnt 6226kj heat was evolved.then the volume of o2(at NTP)that entered in to the rxn is? 1》296.5ml 2》296.5 litre 3》6226×22.4 litre4》22.4 litre.plz explain with solition? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about If heat of combustion of ethylene is 1411kj when a certain amount of ethylene was burnt 6226kj heat was evolved.then the volume of o2(at NTP)that entered in to the rxn is? 1》296.5ml 2》296.5 litre 3》6226×22.4 litre4》22.4 litre.plz explain with solition? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If heat of combustion of ethylene is 1411kj when a certain amount of ethylene was burnt 6226kj heat was evolved.then the volume of o2(at NTP)that entered in to the rxn is? 1》296.5ml 2》296.5 litre 3》6226×22.4 litre4》22.4 litre.plz explain with solition?.
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