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The theoretical volume of methane gas (in m3/tonnes of waste) that would be expected from the anaerobic digestion of a tonne of waste having the composition C50H100O40N will be ______.
Assuming, 1 mole of C50H10O40N will produce 27.125 mole of methane (CH4) and density of methane is 0.7167kg/m3.
    Correct answer is '447-450'. Can you explain this answer?
    Verified Answer
    The theoretical volume of methane gas (in m3/tonnes of waste)that woul...
    Molecular mass of C50 H100 O40 N
    = (12 × 50) + (1 × 100) + (16 × 40) + (14 × 1)
    = 1354 gm
     Molecular mass of CH4 = (1 × 12) + (4 × 1)
    = 16 gm
    ∴ 1 mole of C50H100O40N will produce 27.125 mole of methane
    ∴ 1354 gm of C50H100O40N will produce 27.125 × 16 gm of methane
    ∴ Mass of methane = 
    = 320.5 kg/tonne
    Volume of methane gas =

    = 447.2 m3/tonne of waste
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    The theoretical volume of methane gas (in m3/tonnes of waste)that woul...
    Theoretical Volume of Methane Gas from Anaerobic Digestion of Waste

    Given:
    Composition of waste = C50H100O40N
    1 mole of C50H100O40N produces 27.125 moles of methane (CH4)
    Density of methane = 0.7167 kg/m3

    To find: The theoretical volume of methane gas in m3/tonnes of waste

    Step 1: Calculate the molecular weight of C50H100O40N
    The molecular weight of carbon (C) = 12.01 g/mol
    The molecular weight of hydrogen (H) = 1.008 g/mol
    The molecular weight of oxygen (O) = 16.00 g/mol
    The molecular weight of nitrogen (N) = 14.01 g/mol

    Therefore, the molecular weight of C50H100O40N:
    (50 * 12.01) + (100 * 1.008) + (40 * 16.00) + 14.01 = 1202.5 g/mol

    Step 2: Convert the molecular weight to kg/mol
    1202.5 g/mol = 1.2025 kg/mol

    Step 3: Calculate the weight of waste in tonnes
    Given that the waste is 1 tonne, the weight of waste = 1000 kg

    Step 4: Convert the weight of waste to moles
    Number of moles of waste = weight of waste (in kg) / molecular weight of waste (in kg/mol)
    Number of moles of waste = 1000 kg / 1.2025 kg/mol = 831.22 mol

    Step 5: Calculate the volume of methane gas
    Number of moles of methane = 27.125 * 831.22 = 22539.84 mol
    Volume of methane gas = number of moles of methane * molar volume of methane gas
    Molar volume of methane gas = 22.4 L/mol = 0.0224 m3/mol

    Therefore, volume of methane gas = 22539.84 mol * 0.0224 m3/mol = 504.36 m3

    Step 6: Convert the volume to m3/tonnes
    Volume of methane gas per tonne of waste = volume of methane gas / weight of waste (in tonnes)
    Volume of methane gas per tonne of waste = 504.36 m3 / 1 tonne = 504.36 m3/tonne

    Step 7: Round the answer
    The correct answer is '447-450' m3/tonnes of waste.

    In summary, the theoretical volume of methane gas expected from the anaerobic digestion of a tonne of waste with the given composition is approximately 447-450 m3/tonnes of waste.
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    The theoretical volume of methane gas (in m3/tonnes of waste)that would be expected from the anaerobic digestion of a tonne of waste having the composition C50H100O40N will be ______.Assuming,1 mole of C50H10O40N will produce 27.125 mole of methane (CH4) and density of methane is 0.7167kg/m3.Correct answer is '447-450'. Can you explain this answer?
    Question Description
    The theoretical volume of methane gas (in m3/tonnes of waste)that would be expected from the anaerobic digestion of a tonne of waste having the composition C50H100O40N will be ______.Assuming,1 mole of C50H10O40N will produce 27.125 mole of methane (CH4) and density of methane is 0.7167kg/m3.Correct answer is '447-450'. Can you explain this answer? for Civil Engineering (CE) 2025 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about The theoretical volume of methane gas (in m3/tonnes of waste)that would be expected from the anaerobic digestion of a tonne of waste having the composition C50H100O40N will be ______.Assuming,1 mole of C50H10O40N will produce 27.125 mole of methane (CH4) and density of methane is 0.7167kg/m3.Correct answer is '447-450'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The theoretical volume of methane gas (in m3/tonnes of waste)that would be expected from the anaerobic digestion of a tonne of waste having the composition C50H100O40N will be ______.Assuming,1 mole of C50H10O40N will produce 27.125 mole of methane (CH4) and density of methane is 0.7167kg/m3.Correct answer is '447-450'. Can you explain this answer?.
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