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Chemistry...part 2

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Regarding the question I posted earlier (How many mL of an impure liquid CS2 mixture (density 1.26 gmL-1) which contains 93.0% of CS2 by mass, must be burned to obtain 12.8g SO2?)>>volume is 6.49 mL

what volume of pure oxygen at 25.0 degrees C and 100 kPa would be consumed (assume the carbon content burns to CO2).


  • Chemistry...part 2 -

    Use PV = nRT
    Remember T is Kelvin.
    Use the mass CS2 and convert to moles for n. There are 101.325 kPa in 1 atm.

  • Chemistry...part 2 -

    I can't seem to get the answer (7.43)

    100V= 0.107334995 (8.314) (298.15)

    V= 2.66

    Any ideas?

  • Chemistry...part 2 -

    Rule #1 before we talk about where you went wrong. Don't copy all those digits on the calculator. Use 0.1073 if you wish but no more.
    Now for the problem.
    How did you get the 0.1073 moles.
    I would have started with the 12.8 g SO2 which is 12.8/64 = 0.200 moles SO2. Now convert that to moles O2 (which you didn't do) using the coefficients in the balanced equation.
    0.200 moles SO2 x (3 moles O2/2 moles SO2) = 0.300. Now plug that in. I get 7.436 liters which rounds to 7.44 L. I get 7.43 if I use 298 instead of 298.15 for T and 298.15 isn't justified.

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