The basic reaction involved in the precipitation of phosphorus with aluminium is

Question

The basic reaction involved in the precipitation of phosphorus with aluminium is: Liquid alum [Al2(SO4)3-18H2O] contains 48 % aluminium and based on laboratory testing, 1.5 moles of aluminium react with one mole of phosphorous (P). Given a liquid alum of density 1.2 kg/L and a flow rate of 12 000 m3/day, determine the: a) Amount of liquid alum required to precipitate phosphorus in wastewater that contains 8 mg (10) P/liter. b) Required alum storage capacity if a 30 day supply is to be stored at the treatment facility. (5)

Explanation / Answer

first solving the part a of the problem

a) basis - 1 litre of water

so 1 litre of water contains 8 mg of phosphorous

molecular weight of phosphorous is 30.97g/mole

moles of phosphorous in 1 litre of water is? .008/30.97 = .000258mole

so moles of aluminium required is 1.5 * .000258 =.000387 moles of aluminium is required

so moles of alum = (1/.48)*moles of aluminium (since aluminium contains 48% of aluminum )

= (1/.48)*.000387=.00080625 moles of alum required

mass of alum = moles * ,molecular weight

= .00080625*474.384= .3824g

so we require .3824g of alum / litre of water

b) mass flow rate = volumetric rate * density

voluemetric rate = 12000m^3/day

= 12000*1000liter/day ( since 1m^3=1000litre)

no. of days = 30days

so total mass flow rate =12000*1000*1.2*30

=432000000 kg of storage needed of alum

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.