Question

In: Chemistry

A surface water is coagulated with a dosage of 30 mg/L ferrous sulfate and an equivalent...

A surface water is coagulated with a dosage of 30 mg/L ferrous sulfate and an equivalent dosage of hydrated lime (Ca(OH)2).

2FeSO₄*7H₂O + 2Ca(OH)₂ + 0.5O₂ -> 2Fe(OH)₃ + 2CaSO₄ + 13 H₂O

A ) How many pounds of ferrous sulfate are needed per MGD of water treated.

B) How many pounds/mil of hydrated lime are required assuming, a purity of 70% CaO?

C) Determine the pounds of Fe(OH)3 sludge are produced per MGD of water treated.

Expert Solution

30 mg of FeSO4.7H2O present in 1 liter solution

1 gallon = 3.785 Liters

1 MGD = 3.785 million Liters

1 MGD contains iron sulphate = 3.785 * 30 *10^-3 *10^6 g

= 113550 g

1 g = 0.00220462 pounds

113550 g = 0.00220462 * 113550 pounds

= 250.3346 pounds.

equivalent dosage of Ca(OH)2 is also involved in the reaction.

hence mass of Ca(OH)2 = 250.3346 pounds

the compound is 70 % of Ca(OH)2 = 250.3346 pounds

hence required mass of Ca(OH)2 = 250.3346 * (100/70)

= 357.621 pounds of Ca(OH)2 required.

1 mole of FeSO4 *7H2O produces 1 mole of Fe(OH)3

278 g of FeSO4 *7H2O produces 107 g of Fe(OH)3

250.3346 pounds of FeSO4 *7H2O produces (107/278) * 250.3346 pounds of Fe(OH)3

= 96.352 pounds

queen_honey_blossom answered 2 years ago

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