STP: What volume of gas (in L) is occupied by 16.0g of CH4 at STP?

Is the answer: 22.3L? I need to validate that I've completed the problem correctly. I'm not sure if I am using the appropriate formula! THANKS!

4 Answers

  • 5 years ago
    Favorite Answer


    the molar volume of a gas at STP is 22.4 L/mol

    so convert mass to moles then convert moles to volume at STP

    16.0 g CH4 * (1 mol / 16.0 g CH4) * (22.4 L/mol) = 22.4 L

    not sure where you got 22.3

    maybe you rounded incorrectly

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  • 5 years ago

    To demonstrate WHY 1 mole of any gas has a volume of 22.4 L at STP, use the ideal gas law:

    PV = n RT

    (1 atm) V = 1 mol (0.0821 Latm/molK) (273 K)

    V = 22.4 L

    In your question, as others have shown, 16.0 g CH4 = 1 mol CH4.

  • 5 years ago

    1 mole of any gas occupies 22.4 L in STP,

    V=(16/M) * 22.4

    M is the molecular weight

  • 5 years ago

    1 mole occupies 22.4 L at stp

    as one decimal place is given then 16.0g (16.04g) is still one mole.

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