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    What volume of Cl2(g) at STP would you need to obtain a 250g

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    1. What volume of Cl2(g) at STP would you need to obtain a 250g sample of Cl2(g)?

    2. A .418g sample of gas has a volume of 115mL at 66.3 degrees Celsius and 743mmHg. What is the molar mass of this gas?

    3. What is the volume, in liters, occupied by a mixture of 15.2g Ne(g) and 34.8g Ar(g) at 7.15 atm pressure and 26.7 degrees Celsius?

    4. An 89.3mL sample of "wet" 02(g) is collected over water at 21.3 degrees Celsius at a barometric pressure of 756mmHg (vapor pressure of water at 21.3 degree Celsius = 19mmHg)
    a. What is the partial pressure of 02(g) in millimeters of mercury in the sample collected.
    b. What is the volume percent O2 in the gas collected?
    c. How many grams of O2 are present in the sample?

    5. Calculate the volume of H2(g) measured at 26 degrees Celsius and 751 Torr required to react with 28.5 L CO(g), measured at 0 degree Celsius and 760 torr in the following reaction.
    3 CO(g) + 7 H2(g) --- C3H8(g) + 3H20(l)

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    https://brainmass.com/chemistry/gas-stoichiometry/what-volume-of-cl2-g-at-stp-would-you-need-to-obtain-a-250g-58548

    Solution Preview

    1. Assume Cl2 follow ideal gas law:
    PV=nRT
    At STP conditions:
    T=273K
    P=1atm
    R=0.082 Latm/molK

    So you need the number of mole of Cl2
    Molecular weight of Cl2=2*35.5=71(g/mol)
    Number of mole of Cl2=250/71=3.52(mol)

    Substitute into PV=nRT
    V=nRT/P=3.52*0.082*273/1=78.8(L)

    Reference:
    http://www.katmarsoftware.com/gconvals.htm

    2. Using PV=nRT
    P=743/760=0.98(atm)
    R=0.082 Latm/molK
    V=0.115(L)
    T=66.3+273.15=339.45(K)

    We can calculate ...

    Solution Summary

    The solution provides detailed explanations and calculations for the problem.

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