What is the volume of dichromate ion solution required in the following scenario:
Ionic Equation : 3SO2+ Cr2O7 (2-) + 2H (+) ----> 3SO4 (2-) + 2Cr(3+) + H20
A solution contains 0.1 mol/dm^3 of dichromate (VI) ions. Calculate the minimum volume of this solution required to removed the sulfur dioxide from 3.0 dm^3 of polluted air which contains 4.8 % by volume of sulfur dioxide measured at s.t.p
The ionic equation given here is 3SO2 + Cr2O7 (2- ) + 2H (+) --> 3SO4 (2- ) + 2Cr (3+) + H20.
The polluted air contains an amount of SO2 that occupies 4.8% of 3.0 dm^3 of the air. For every three molecules of SO2 we require 1 ion of Cr2O7 (2- ).
At STP, using PV = nRT, the moles of SO2 in the air is given by:
n = PV / RT
=> n = [4.8%*100*1000*(3/1000)]/ 273*8.3145
=> n = 4.8*3/273*8.3145
=> n = 6.34 * 10^-3 mol.
So we need the dichromate ions for 6.34 * 10^-3 moles of SO2.
We need 2.11* 10^-3 mol of dichromate ion.
Now there is 1 mole of dichromate ions in 10 dm^3 of solution or in 10/1000 m^3 of the solution.
2.11*10^-3 moles will be found in 2.11*10^-3*10^-2 m^3 of the solution.
The minimum volume of dichromate solution required is 2.11*10^-5 m^3.