How many grams of ethane gas (C2H6) are in a 12.7 liter sample at 1.6 atmospheres and 24°C? Show all work used?

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  • Jim
    Lv 7
    9 years ago
    Favorite Answer

    28.05 Grams per Mole

    PV = nRT

    K = °C + 273.15, therefore 24C = 24 + 273.15 = 297.15K

    1 atm = 101325 Pa = 101.325 kPa, therefore, 1.6 ATM = 1.6 x 101.325 kPa = 162.12 kPa

    P=pressure

    V=volume = 12.7L

    n=moles of gas

    T=temperature = 297.15K

    R = gas constant (dependent on the units of pressure, temperature and volume)

    R = 8.314 J K-1 mol-1 if

    Pressure is in kilopascals(kPa) = 162.12 kPa

    Volume is in litres(L)

    Temperature is in kelvin(K)

    R = 0.0821 L atm K-1 mol-1 if

    Pressure is in atmospheres(atm)

    Volume is in litres(L)

    Temperature is in kelvinKelvin(K)

    Solve the equation for n, number of moles of gas:

    PV/RT = n

    n = (162.12 kPa)(12.7L)/(8.314 J K-1 mol-1)(297.15K)

    n = 2058.924/2470.5051

    n = 0.8334 Moles

    And since 1 mole of Ethane has a MW of 28.05 Grams, then

    Grams of Ethane = 0.8334 x 28.05 = 23.38 Grams, approximately

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