Question

Ferrous sulfate is added along with lime (CaO) to water to raise the pH and form ferric hydroxide particles to aid coagulation. Consider the reaction to be as follows:
2 FeSO4·7H2O + 2 CaO + 0.5 O2 → 2 Fe(OH)3(s) + 2 CaSO4 + 4 H2O
The ferrous sulfate dose is 40 mg/L.

THE MOST IMPORTANT PARTS TO ANSWER ARE b and d. Please explain clearly and show ALL working. Its very useful to help me understand the problem. THANKS!!!
(a) Determine how many pounds of ferrous sulfate are required for every million gallons of water treated.
(b) Determine how many pounds of lime are required for every million gallons of water treated.
(c) Determine how many pounds (dry weight) of ferric hydroxide solids are theoretically produced per million gallons of water treated.
(d) If the ferric hydroxide solids are collected in a clarifier as a 2% by weight slurry of solids in water, and that slurry is then put through a belt filter press to dewater the slurry to 25% by
weight solids, determine how many pounds of ferric hydroxide sludge (solids plus water) will be produced from the belt filter press per million gallons of water treated.

Answer 1

Answer a) ferrous sulphate dose = 40 mg/l

For 1000000 gallons , ferrous sulphate = 40*3.785*1000000 mg = 151.4*1000000/1000000 kg = 151.4 kg = 151.4* 2.205 = 333.837 lb

Answer b) 2 moles of ferrous sulphate require 2 moles of lime moles of ferrous sulphate = 151.4/ molecular weight of ferrous sulphate = 151.4*1000/278 thus lime required = 151.4*56/278 = 30.497 kg = 30.497*2.205 lb = 67.246 lb

Answer c) Pounds of Fe(OH)3 produced = 151.4*107*2.205 /278 = 128.49 lb

Answer d) Fe(OH)3 ions in clarifier : weight of slurry = 6424.5 lb and weight of water = 6296.01 lb

Weight of Fe(OH)3 in belt filter press = 128.49 lb , weight of slurry = 128.49*100/25 = 513.96 lb

Thus weight of solids plus water in belt filter press per million gallons of water treated = 513.96 lb