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11 July, 05:27

How many grams of methane gas (CH4) need to be combusted to produce 12.5 L water vapor at 301 K and 1.1 atm? Show all of the work used to solve this problem.

CH4 (g) + 2O2 (g) yields CO2 (g) + 2H2O (g)

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  1. 11 July, 08:50
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    Answer is: 4.45 grams of methane gas need to be combusted.

    Balanced chemical reaction: CH₄ + 2O₂ → CO₂ + 2H₂O.

    Ideal gas law: p·V = n·R·T.

    p = 1.1 atm.

    T = 301 K.

    V (H ₂O) = 12.5 L.

    R = 0,08206 L·atm/mol·K.

    n (H₂O) = 1.1 atm · 12.5 L : 0,08206 L·atm/mol·K · 301 K.

    n (H₂O) = 0.556 mol.

    From chemical reaction: n (H₂O) : n (CH₄) = 2 : 1.

    n (CH₄) = 0.556 mol : 2 = 0.278 mol.

    m (CH₄) = 0.278 mol · 16 g/mol.

    m (CH₄) = 4.448 g.
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