Forty grams of CH4 gas at STP occupies what volume?

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  • 1 decade ago
    Favorite Answer

    PV = nRT

    P= 1 atm

    R=.0821

    T=273 K

    n=weight / mol.mass = 40/16

    therefore, V = nRT/P

    V = (40*0.0821*273)/(16*1)

    = 56 Litres

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  • Anonymous
    1 decade ago

    The molar mass of CH4 is 16 grams per mole. One mole of any gas at STP occupies 22.4 Liters. Since you have 40 grams you divide that by the 16 grams and get the number of Liters of gas. 40/16 = 20/8 = 10/4 = 2.5 Liters.

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  • Anonymous
    1 decade ago

    This question can be answered using the ideal gas equation PV=nRT

    n=no. of moles=mass of the substance/molar mass=40/12+4=40/16=2.5 moles. V=nRT/P=2.5 x 0.0821 x 273.15K/0.987atm=56.8 liters. R=gas constant

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  • 1 decade ago

    Molar mass CH4 = 12 + 4 =16

    No. of moles in 40g = 40/16 = 2.5moles

    PV = nRT

    101.3 V = 2.5*8.31*273

    V = 56.0 litres

    Chech 1mol at STP = 22.4 litres

    2.5 moles at STP = 2.5*22.4 = 56.0litres

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  • 1 decade ago

    If you are using the ideal gas law, use pV=nRT where p=pressure, V=volume, n=number of moles, R=universal gas constant, T=temperature.

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  • 1 decade ago

    CH4=16g/mole ; 1mole=22.4L/mole@stp

    40g/16g/molex22.4L/mole=56.0L@stp

    Source(s): chem major
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  • 4 years ago

    40(mass) / 14.01 (molar mass) = 2.855 moles 2.855 x 22.4 = 63.95 L

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