Question

+60t 13th using the following reaction 2 -> CaOH(pan (5) 5 Ca (OHly +3 HP What would be the daily volume of Ca coule solution of Concentration 8.5916 (as (a) required to treat PH adrested waste water containly steps 3.4 mg/l Phosphorous? the treatment plant process 10000 m of water per day A Approx. 2.2 x103 L B) Approx 3.8x103 L C) Approx 5.48103 L D) Approx. 8.6*103 L El Approx, 9.7 X3 L

Answer 1

Amount of phosphorus in 1L water = 3.4 mg

Total volume of water = 10000 cubic m = 10000000 L

Amount of phosphrus in total water = 34000 g

Moles of phosphorus = Given Wt / Molecular Wt

= 34000 g/ 96 g/mol

= 354.2 mol

Now,

As per the balanced equation 3 mol HPO4 (2-) reacts with 5 mol of Ca(OH)2

3 mol HPO4 (2-) : 5 mol Ca(OH)2

354.2 mol : 590 mol Ca(OH)2

The Ca(0H)2 solution contains 8.5 g Ca(OH)2 in 1L solution.

Moles of Ca(OH)2 = 8.5 g/74g/mol = 0.115 mol

THus, 0.115 mol are present in 1L of solution.

Hence,

Molarity of solution = 0.115 M

Molarity of solution = Moles / Volume (L)

So,

0.115 M = 590 mol / Volume

Volume = 5130.5 L

Correct option : C