Alex asked in Science & MathematicsChemistry · 1 decade ago

what is the volume of 79.3 g of Cl2 at STP?

4 Answers

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  • Lexi R
    Lv 7
    1 decade ago
    Favorite Answer

    At STP 1 mole of any gas occupies a volume of 22.4 L

    moles Cl2 = mass / molar mass

    = 79.3 g / 70.9 g/mol

    = 1.12 mol

    Volume = 1.12 mol x 22.4 L/mol

    = 25.1 L (3 sig fig)

    PV = nRT should be used for any conditions other then STP (which is 1 atm and 0 deg C (273.15 K) It can be used for STP as well, but since we know volume of 1 mole ideal gas = 22.4 L at STP it is not essential.

  • 1 decade ago

    1 mol of Cl2 / 70.9 g of Cl2

  • Vix
    Lv 5
    1 decade ago

    1 mol of Cl2 = 70.9 g

    1 mol of Cl2 = 22.4 mol

    73.9 g of Cl2 * 1 mol of Cl2 / 70.9 g of Cl2 * 22.4 L of Cl2/1 mol of Cl2=

    23.3 L of Cl2

  • MacD
    Lv 4
    1 decade ago

    use the ideal gas equation.

    PV = nRT

    V = (nRT)/p

    STP= T = 273.15 K ; P = 1 atm.

    n= mass/ molar mass

    = 79.3/35.45= 2.237moles

    V = [(2.237moles)(0.082057)(273.15 K)] / 1 atm

    =50.14 L(liters)

    i don't know where you had that formula vix but it is wrong.

    Source(s): chemistry undergrad
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