Page 211 - Materials Chemistry, Second Edition

194 Practical Design Calculations for Groundwater and Soil Remediation

• Soil porosity = 0.35
• Formula of gasoline (assumed) = C H 16
7
• The amounts of N and P naturally occurring in the excavated soil
are insignificant
• Trisodium phosphate (Na PO ⋅12H O) as the P source; price = $10/lb
4
3
2
• Ammonium sulfate ((NH ) SO ) as the N source; price = $3/lb
4 2
4
• One-time nutrient addition only.
Solution:
(a) Determine the number of moles of gasoline:

MW of gasoline = 7×12 + 1×16 = 100
Moles of gasoline = 158/100 = 1.58 kg-mole

(b) Determine the number of moles of C in soil:
Since there are seven carbon atoms in each gasoline molecule, as
indicated by its formula, C H , then,
16
7
Moles of C = (1.58)(7) = 11.06 kg-mole

(c) Determine the number of moles of N needed (using the C:N:P
ratio):

Moles of N needed = (10/100)(11.06) = 1.106
kg-mole
Moles of (NH ) SO needed = (1.106)/2
4
4 2
= 0.553 kg-mole (each mole of
ammonium sulfate contains
two moles of N)
Amount of (NH ) SO needed = (0.553)[(14+4)(2) + 32 + (16)(4)]
4
4 2
= 73 kg = 161 lb = $483
(d) Determine the number of moles of P needed (using the C:N:P
ratio):

Moles of P needed = (1/100)(11.06) = 0.111 kg-mole
Moles of Na PO ∙12H O needed = 0.111kg-mole
4
2
3
Amount of Na PO ∙12H O needed
3
2
4
= (0.111)[(23)(3) + 31 + (16)(4)+ (12)(18)]
= 42 kg = 92.5 lb = $925