Page 45 - Materials Chemistry, Second Edition

28 Practical Design Calculations for Groundwater and Soil Remediation

= [(110 yd )(27 ft /yd )]{(1.65 g/cm )[62.4 lb/ft /(1 g/cm )]}
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= (2,970)(103) lb = 305,900 lb = 139,000 kg
(b) Mass of COC in soil = (mass of soil)(COC concentration in soil)
Mass of TPH = (139,000 kg)(1,000 mg/kg)
= 1.39 × 10 mg = 1.39 × 10 g
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Mass of benzene = (139,000 kg)(20 mg/kg) = 2.78 × 10 mg
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= 2.78 × 10 g
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Mass of toluene = (139,000 kg)(20 mg/kg) = 2.78 × 10 mg
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= 2.78 × 10 g
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Mass of ethyl benzene = (139,000 kg)(20 mg/kg) = 2.78 × 10 mg
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= 2.78 × 10 g
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Mass of xylenes = (139,000 kg)(20 mg/kg) = 2.78 × 10 mg
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= 2.78 × 10 g
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(c) Mass fraction of a COC = (mass of the COC) ÷ (mass of TPH)
Mass fraction of benzene = (2.78 × 10 )/(1.39 × 10 ) = 0.020 = 2.0%
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Mass fraction of toluene = (2.78 × 10 )/(1.39 × 10 ) = 0.020 = 2.0%
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Mass fraction of ethyl benzene = (2.78 × 10 )/(1.39 × 10 ) = 0.020
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= 2.0%
Mass fraction of xylenes = (2.78 × 10 )/(1.39 × 10 ) = 0.020 = 2.0%
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(d) Moles of a COC = (mass of the COC) ÷ (MW of the COC)
Moles of TPH = (1.39 × 10 )/(100) = 1,390 g-mole
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Moles of benzene = (2.78 × 10 )/(78) = 35.6 g-mole
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Moles of toluene = (2.78 × 10 )/(92) = 30.2 g-mole
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Moles of ethyl benzene = (2.78 × 10 )/(106) = 26.2 g-mole
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Moles of xylenes = (2.78 × 10 )/(106) = 26.2 g-mole
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(e) Mole fraction of a COC
= (moles of the COC)/(moles of TPH)
Mole fraction of benzene = (35.6)/(1,390) = 0.0256 = 2.6%
Mole fraction of toluene = (30.2)/(1,390) = 0.0217 = 2.2%
Mole fraction of ethyl benzene = (26.2)/(1,390) = 0.0189 = 1.9%
Mole fraction of xylenes = (26.2)/(1,390) = 0.0189 = 1.9%