posted by Matt .
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
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.
gm^-3 = g/m^3
moldm^-3 = mol/dm^3 = mols/L = M
dm^3 = L
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
You could have just posted a solution.