Dichloroethane C 2 H 4 Cl 2 is a common industrial solvent. underground behind a

Question

Dichloroethane C2H4Cl2 is a common industrial solvent. underground behind an abandoned plant a 700m3 volume of soil is found to be contaminated with this chemical to an average concentration of 50 ppm. a survey of the soil indicates that ample bacteria are present to decompose the waste according to the following equation:

2C2H4Cl2 + 5O2 ---> 4CO2 + 4HCl + 2H2O

Presently, however, there is insufficient oxygen available to the microorganisms, and therefore you decide to pump water containing 8 ppm O2 into the site at a rate of 50L/min. Assume that both the bacteria and dichloroethane are adsorbed irreversibly to the soil. Also assume that 10% of the supplied oxygen is used by bacteria to degrade dichloroethane according to the above equation, while the remaining 90% wasted to the soil or to the other microbial processes. How much time is required for the average concentration of C2H4Cl2 at the site to fall to 10 ppm?

Explanation / Answer

Given the concentration of dichloroethane = 50 ppm = 50 mg/L

Total volume of contaminated soil = 700 m3 = 700 m3 x (1000L / 1m3) = 700000 L

The mass of Dichloroethane need to be decomposed in order to decrease the concentration to 10 ppm

= (50 - 10) mg /L x 700000 L = 2.8 x 107 mg = 2.8 x 104 g Dichloroethane

Molecular mass of Dichloroethane(C2H4Cl2) = 98.96 g/mol

Hence moles of Dichloroethane(C2H4Cl2) need to be decomposed = 2.8 x 104 g / 98.96 g/mol = 282.94 mol

The balanced chemical reaction is

2C2H4Cl2 + 5O2 ---> 4CO2 + 4HCl + 2H2O

2 mol, ------- 5mol ----- 4 mol, 4 mol, 2 mol

2 mol of C2H4Cl2 reacts with 5 mol of O2

Hence  282.94 mol of C2H4Cl2 that will react with the moles of O2

= 282.94 mol C2H4Cl2 x ( 5 mol O2 / 2 mol C2H4Cl2) = 707.3565 mol O2

Hence mass of O2 (MW = 32 g/mol) required = 707.3565 mol O2 x 32.0 g/mol = 22635.41 g O2

Only 10 % of the supplied O2 is used to decompose C2H4Cl2.

Hence mass of O2 need to be supplied = 22635.41 g O2 x (100 / 10) = 226354.1 g O2 = 226354100 mg O2

Given the concentration of O2 in water = 8 ppm = 8 mg/L

Hence volume of water containing  226354100 mg O2 = 226354100 mg / 8 mg/L = 28294262.5 L

Given the rate of flow of water = 50 L / min

Hence the total time required to supply 28294262.5 L water = 28294262.5 L / 50 L / min = 565885 min

= 565885 min x (1 hr / 60 min) x (1 day / 24 hr) = 393 days (answer)

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