Ozone, O_3, is an important species in the chain of reactions that lead to the p

Question

Ozone, O_3, is an important species in the chain of reactions that lead to the production of smog. In an ozone analysis 2.0 Times 10^5 of air at STP were drawn through a solution of Nal where the O_3 undergoes a reaction, O_3 + 2I^- + H_2 O rightarrow O_2 + I_2 + 2OH^- The 12 formed was titrated with 0.0100 M Na_2S_2O_2 with which it reacts. I_2 + 2 S_2O_3^2+ Rightarrow 2I^- + S_4O62+_6 In the analysis, 0.420 ml of the Na_2S_2O_3 solution was required to completely react with all of the I_2. Calculate the number of moles of I_2 that were reacted with the Na_2S_2O_3 solution. How many moles of I_2 were produced in the first reaction? How many moles of O_3 were contained in the 2.0 Times 10^5 L of air? How many milliliters would the O_3 occupy at STP? What is the concentration of O_3, in parts per million by volume, in the air sample ?

Explanation / Answer

1)moles of I2=1/2 moles of Na2S2O3=1/2*0.01 mol/L*0.420 ml*1L/1000ml=2.1*10^-6 moles

2) 2.1*10^-6 moles

3)moles of I2=moles of O3 (from the first rxn)= 2.1*10^-6 moles

4)At STP,1 mole of any ideal gas=22.4 L

So volume of O3=22.4L/mol*2.1*10^-6 moles=47.04*10^-6 L=4.70*10^-5 L=4.70*10^-5 L*1000 ml/L=4.70*10^-2 ml

5)O3 per volume of air=(4.70*10^-5 L/2.0*10^5L)*10^-6=2.35*10^-16 ppm

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