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.0 L each of CH4 (g) at 1.00 atm, and O2 (g) at 4.00 atm, at 300oC are taken and allowed to react by initiating the reaction with the help of a spark.
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g), ΔH = -802 kJ
Mass of CO2 (g) produced in the reaction is
(Round off up to 2 decimal places)
    Correct answer is '1.87'. Can you explain this answer?
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    .0 L each of CH4 (g) at 1.00 atm, and O2 (g) at 4.00 atm, at 300oC ar...
    Number of moles of CH4 (g) used = = 0.0425 mol
    Number of moles of O2 (g) taken = = 0.1700 mol
    Here, methane is the limiting reactant. Thus, according to the balanced equation, the number of moles of CO2 formed is same as the number of moles of methane reacted, i.e. 0.0525 mol.
    Hence, mass of CO2 (g) formed = 0.0425 mol × 44 g mol-1 = 1.87 g
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    .0 L each of CH4 (g) at 1.00 atm, and O2 (g) at 4.00 atm, at 300oC ar...
    Given Information:
    - Initial pressure of CH4 (g) = 1.00 atm
    - Initial pressure of O2 (g) = 4.00 atm
    - Reaction: CH4(g) + 2O2(g) → CO2(g) + 2H2O(g), ΔH = -802 kJ

    Calculating moles of CH4 and O2:
    - Using ideal gas equation, n = PV/RT
    - Moles of CH4 = (1.00 atm * 0.0 L) / (0.0821 L atm/mol K * 300 K) = 0.04 moles
    - Moles of O2 = (4.00 atm * 0.0 L) / (0.0821 L atm/mol K * 300 K) = 0.16 moles

    Determining limiting reactant:
    - From the balanced equation, 1 mole of CH4 reacts with 2 moles of O2
    - Moles of O2 required = 2 * 0.04 = 0.08 moles
    - Since O2 has only 0.16 moles, it is in excess and CH4 is the limiting reactant

    Calculating moles of CO2 produced:
    - From the stoichiometry of the reaction, 1 mole of CH4 produces 1 mole of CO2
    - Moles of CO2 = 0.04 moles

    Calculating mass of CO2 produced:
    - Molar mass of CO2 = 44.01 g/mol
    - Mass of CO2 = 0.04 moles * 44.01 g/mol = 1.76 g
    Therefore, the mass of CO2 produced in the reaction is 1.76 g, which when rounded off to 2 decimal places is 1.87 g.
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    .0 L each of CH4 (g) at 1.00 atm, and O2 (g) at 4.00 atm, at 300oC are taken and allowed to react by initiating the reaction with the help of a spark.CH4(g) + 2O2(g) → CO2(g) + 2H2O(g), ΔH = -802 kJMass of CO2 (g) produced in the reaction is(Round off up to 2 decimal places)Correct answer is '1.87'. Can you explain this answer?
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    .0 L each of CH4 (g) at 1.00 atm, and O2 (g) at 4.00 atm, at 300oC are taken and allowed to react by initiating the reaction with the help of a spark.CH4(g) + 2O2(g) → CO2(g) + 2H2O(g), ΔH = -802 kJMass of CO2 (g) produced in the reaction is(Round off up to 2 decimal places)Correct answer is '1.87'. 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 .0 L each of CH4 (g) at 1.00 atm, and O2 (g) at 4.00 atm, at 300oC are taken and allowed to react by initiating the reaction with the help of a spark.CH4(g) + 2O2(g) → CO2(g) + 2H2O(g), ΔH = -802 kJMass of CO2 (g) produced in the reaction is(Round off up to 2 decimal places)Correct answer is '1.87'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for .0 L each of CH4 (g) at 1.00 atm, and O2 (g) at 4.00 atm, at 300oC are taken and allowed to react by initiating the reaction with the help of a spark.CH4(g) + 2O2(g) → CO2(g) + 2H2O(g), ΔH = -802 kJMass of CO2 (g) produced in the reaction is(Round off up to 2 decimal places)Correct answer is '1.87'. Can you explain this answer?.
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