Question

organic contaminants may be represented by the generic molecular formula CmHnOpNr where m,n,p and r represent integers. the oxidation of an organic may be carried out in two step, carbonaceous oxidation followed by nitrogenous oxidation. determine the value of the stoichiometric coefficient a, b b and d in term of m, n, p and r. give an expressiOn for the theoretical oxygen demand per gram of organic due to carbonaceous oxidation alone (give the answer in unit of gram O₂ per gram of organic) Give an expression for the total theoretical oxygen demand (carbonaceous and nitrogenous) per gram of organic Try out your new expression by computing both the carbonaceous and the nitrogenous theoretical oxygen demand associated with 150 mg/L glycine C₂H₅O₂N.

CmHnOpNr + aO2 bCO2 + cH₂O + dNH₃

NH₃ + 2O₂ NO₃^- + H₂O + H⁺

Answer 1

Cụ , , + 4 , -> bo, H C H, O + 4 NH, 10x20x4(2êu4 - oxidation NH3 + 202 NO3 + H2O + Hit 1 Nitrogenous oxidation mab .. na 2C+38 m 2c +30 Dac=n-3r p+ 2a a abtc | c=0.5n-1.50 rad [bim [der p+za=2b+c => p+2a= am to 5n-1.50 » [a= m +0.25n-0.757-0.5p 1 Calbonaceous Oxidation Thod =2a = 2x16x a Molecular weight of organic = 12m+n+laptlar !. Thod 3 320 12m tntioptiur Tonal Thod= 32a (918) +ad (32) (1 ms of Nn3) - 12m intifp+148 12M+n +16+1470 Hyr (olg) of organic D 150 mgle of Clycine C2H5O2N Molecular at = 24+5+32714 - m=2, n=5, p=2, r=1 75g|MOT is as 3+0.25(5) -0.75(1)-0.512) = 1.5 to Carbon accous ThoD = 32x115 g/ R ose :. In given Case Carbon accous ThoD = 32777 x 1890 - Very 75V . 32x1. 5x iso my = 96 mg 3271-5 * 150 75 Nitrogenous Thod= 64d gl mode of NH3 . = 64 X1 X 2 mg = 128 mg or onesny EL 3 x 150 mg = 12eug