Question

A new method for the destruction of a chlorine-containing toxic waste such as a polychlorinated biphenyl (PCB), produces CO2 and HCl, according to the following equation:

C12 H4 + H2O + O2 》 CO2 +HCI

a) Balance the equation
b) A 10.87 kg sample of soil, suspected to be contaminated with PCB, is extracted with diglyme (a solvent that PCB is soluble in) until no organic material remains in the soil. The liquid diglyme is removed by distillation and the remaining material is transferred to the reactor, at which point it is sealed and heated for 24 hours in excess oxygen and water. After reaction, 248.9 mL of 0.01000M NaOH is required to titrate all of the HCl produced in the reaction. Determine the %PCB in the original sample.

Answer 1

polychlorinated biphenyl is represented as C12H_10-x Cl_x

here x =6

C12H4Cl6+H2O+ 11.5 O2 -------->12CO2+ 6HCl

the reaction between HCl and NaOH is HCl+ NaOH------>NaCl+ H2O

1 mole of HCl requires 1 mole of NaOH to get neutralized

moles of NaOH=

molarity* Volume(L)= 0.01*248.9/1000 =0.0025= moles of HCl

6 moles of HCl is produced from 1 mole of polychlorinated biphenyl

0.0025 moles are produced from 0.0025/6= 0.000417

molar mass of C12H4Cl6= 12*12+4+35.5*6= 361

mass of C12H4Cl6= moles* molar mass = 0.000417*361= 0.15 gm

% PCB in the sample = 100*0.15*10^-3 kg/10.87= 0.00138%