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What volume would be occupied by 88.0 g of gaseous carbon monoxide at STP?

1 Answer

  • 9 years ago
    Favorite Answer

    STP is Standard Temperature and Pressure, so you have to use the formula PV=nrt


    P = Pressure = 1 atm

    V = Volume = (unknown)

    n = moles of CO

    R = gas constant = 0.0821 L*atm / Kmol

    T = Temperature = 273K

    First do a mass to mole conversion for carbon monoxide (CO)

    n = weight/molar mas

    weight = 88g (the given mass of CO)

    molar mass = 28 (C =12g/mol, O = 16g/mol : 12 + 16 = 28g/mol)

    n = 88g / 28g/mol

    n = 3.1429 moles

    Now solve for Volume:

    V = nRT/P

    V = (3.1429)(0.0821)(273) / (1)

    V = 70.4418 L

    The volume occupied by 88.0g of CO gas at STP is 70.4 L

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