To remediate one kilogram of dichloromethane (CH2Cl2) using sodium permanganate (NaMnO4) for chemical oxidation, approximately 7.94 kilograms of NaMnO4 would be required. For chemical reduction using zero valent iron (Fe0), approximately 2.63 kilograms of Fe0 would be required.
To remediate one kilogram of dichloromethane (CH2Cl2) using sodium permanganate (NaMnO4) for chemical oxidation, we need to determine the mass of NaMnO4 required.
- Write the balanced chemical equation for the oxidation: 2CH2Cl2 + 9NaMnO4 -> 2CO2 + 10MnO2 + 9NaCl + 4H2O
- Calculate the molar mass of CH2Cl2: 12.01 g/mol (C) + 2(1.01 g/mol) + 2(35.45 g/mol) = 84.93 g/mol
- Convert the mass of CH2Cl2 to moles: 1 kg / 84.93 g/mol = 11.77 mol
- Use the stoichiometry of the balanced equation to find the moles of NaMnO4 needed: 9 mol NaMnO4 / 2 mol CH2Cl2 = 52.97 mol NaMnO4
- Convert the moles of NaMnO4 to grams: 52.97 mol x 149.89 g/mol = 7940 g (rounded to nearest whole gram)
Therefore, approximately 7.94 kilograms of sodium permanganate (NaMnO4) would be required to remediate one kilogram of dichloromethane (CH2Cl2) using chemical oxidation.
For chemical reduction using zero valent iron (Fe0), the balanced chemical equation is: CH2Cl2 + 4Fe0 -> CH4 + 4FeCl2
- Calculate the molar mass of CH2Cl2: 84.93 g/mol
- Convert the mass of CH2Cl2 to moles: 1 kg / 84.93 g/mol = 11.77 mol
- Use the stoichiometry of the balanced equation to find the moles of Fe0 needed: 4 mol Fe0 / 1 mol CH2Cl2 = 47.06 mol Fe0
- Convert the moles of Fe0 to grams: 47.06 mol x 55.85 g/mol = 2630 g (rounded to nearest whole gram)
Therefore, approximately 2.63 kilograms of zero valent iron (Fe0) would be required to remediate one kilogram of dichloromethane (CH2Cl2) using chemical reduction.