Page 222 - Materials Chemistry, Second Edition

Vadose Zone Soil Remediation 205

Solution:
(a) Let us start with oxygen as the oxidant:
C H (CH ) + 10.5O → 8CO +5H O
4
6
3 2
2
2
2
As shown in this equation, the stoichiometric requirement is 10.5
moles of oxygen per mole of xylenes. To express the oxygen
requirement on the basis of g O /g xylenes:
2
MW of C H (CH ) = (12)(8) + (1)(10) = 106
6
3 2
4
Concentration of xylenes = 5,000 mg/kg = 5.0 g/kg soil
= (5.0 g ÷ 106 g/mole)/kg soil = 4.72 × 10 mole xylenes/kg
−2
soil
Stoichiometric amount of O (using the molar ratio)
2
= (10.5 moles of O /mole of xylenes) × (4.72 × 10 mole
−2
2
xylenes/kg soil)
= 0.495 mole O /kg soil
2
= (0.495 mole × 32 g/mole O )/kg soil
2
= 15.85 g O /kg soil
2
Oxygen requirement (in mass ratio)
= (10.5 moles of O /mole of xylenes) × [(32 g/mole) ÷ (106 g/
2
mole)]
= 3.17 g O /g xylenes
2
Stoichiometric amount of O (using the mass ratio)
2
= (3.17 g O /g xylenes) × (5.0 g xylenes/kg soil)
2
= 15.85 g O /kg soil
2
(b) Now let us determine the stoichiometric amount of sodium per-
sulfate, if it is the oxidant to be applied.
MW of sodium persulfate (Na S O ) = (23)(2) + (32)(2) + (16)(8) = 238
2 2
8
As shown in Table 5.3 and as previously discussed, the stoichio-
metric requirement of sodium persulfate will be two times
that of oxygen.
Stoichiometric amount of Na S O (using the molar ratio)
2 2
8
= (2 moles of Na S O /mole of O ) × (0.495 mole O /kg soil)
8
2
2
2 2
= (0.99 mole Na S O /kg soil)
2 2
8
= (0.99 mole × 238 g/mole Na S O /kg soil)
2 2
8
= 236 g Na S O /kg soil
2 2
8