# Chemistry

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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

gm^-3
moldm^-3
dm^3

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.

Molarity=moles/Liters
Solve for liters,

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

• Chemistry - ,

Dr.Bob222,

You could have just posted a solution.

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