ead in drinking waters is 0.015 mg L (or 15 ppm). Before you can need to convert

Question

ead in drinking waters is 0.015 mg L (or 15 ppm). Before you can need to convert this concentration of lead to units of molarity. drinking wat c. The EPA limit for What is 0.015 mg L of lead in units of moles per liter? d. Okay, here's where we figure out if the plan is chemically feasible (based on our assumptions). Would adding the blocks of limestone precipitate enough lead carbonate to lower the levels of lead in the water to below the EPA limit? e. (Optional Bonus Question) The project would cost $5,000 plus $1060.00 per metric ton of purified calcite limestone to be used (a metric to 2 billion liters, what would be the minimum cost of the project? n is 1000 kg). If the volume of the lake is

Explanation / Answer

c. EPA limit of Lead is = 0.015 mg/L

= 0.015/1000 g/mg . mg/L

= 1.5 x 10-5 g/L

Concentration = ( mass / L ) / Molar mass of Pb

= 1.5 x 10-5 g/L / 207.2 mol/g

= 7.24 x 10-8 mol/L

d. 1 mole of CaCO3 would be required to precipitate Pb, but slight heat would be required to break CaCO3 into CaO and CO2. Also by only using limestone could increase the alkalinity of water and hard water can be formed.

e. The mole of Pb to neutralize = 7.24 x 10-8 mol/L

Hence moles of CaCO3 require = 7.24 x 10-8 mol/L

Mass of CaCO3 = No of moles x Molar mass of CaCO3

= 7.24 x 10-8 x 100 g/L

= 7.24 x 10-6 g/L

For 2 billion liter mass of CaCO3 require = 2 x 109 x 7.24 x 10-6 g

= 1.45 x 104 g

= 1.45 x 104 / 106 metric tonn

= 1.45 x 10-2 metric tonn

Cost of CaCO3 = 1.45 x 10-2 metric tonn x 1060 $ / metric tonn = 15.37 $

Total project cost = 5000 + 15.37 = 5015.37 $

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