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Air in the vicinity of a local power station contains 8.8x10^-4 gm^-3 SO2. What volume of 0.001moldm^-3 Iodine solution would be required to react with a solution obtained from 200dm^3 of contaminated air?

The equation given is
SO2 + I2 + 2H2O ==> H2SO4 + 2HI

Thanks! :)

  • Chemistry - ,

    I would help, but im a little bit confused about some of your notations


    Someone else may come along and help, but just in case a clarification would be nice.

  • Chemistry - ,

    gm^-3 = g/m^3
    moldm^-3 = mol/dm^3 = mols/L = M
    dm^3 = L

  • Chemistry - ,

    1 mole of SO2= 1 mole of I2=2 moles of H2O.

    Density=mass/volume, solve for vol.

    Density*volume=mass=(8.8x10^-4 g/m^3)* 220 m^3= g of SO2.

    g of SO2/molecular weight of SO2= moles of SO2

    Since the reaction requires 1 mole of SO2 to react with 1 mole of I2, the moles that were calculated for SO2 must equal to moles of I2.

    Solve for liters,

    Liters of I2 required=moles/Molarity=moles of SO2/0.001 of I2

  • Chemistry - ,


    You could have just posted a solution.

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