Question

A water treatment plant receives the source water with an average Ca2+ concentration of 46.9 mg/L and Mg2+ concentration of 14.8 mg/L. The plant is treating 80 million gallons of water per day. What mass of solids will be produced per day if all of the calcium and magnesium are converted to CaCO3(s) and Mg(OH)2(s) in the softening process? Give your answer in kg.

Answer 1

Answer:

42,650 kg of calcium carbonate will be produced everyday.

13,600.5 kg of magnesium hydroxide will be produced everyday.

Explanation:

Concentration of calcium ions = 46.9 mg/L

Concentration of magnesium ions = 14.8  mg/L

Volume of solution treated everyday , V= 80 million gal

= $80\times 10^6 gal=4.546\times 80\time 10^6 L=3.637\times 10^8 L$

1 gallon = 4.546 Liter

Mass of  calcium ion in V = $46.9 mg/L\times 3.637\times 10^8 L$

=  $1.706\times 10^{10} mg$

1 mg = 0.001 g

$1.706\times 10^{7} g$

Moles of calcium ions = $\frac{1.706\times 10^{7} g}{40 g/mol}=426,500 mol$

From 1 mole of calcium ion 1mol of carbonate is formed . then from 426,500 moles of calcium ion will form :

$\frac{1}{1}\times 426,500 mol=426,500 mol$ of calcium carbonate

Mass of 426,500 moles of calcium carbonate:

426,500 mol × 100 g/mol  = 42,650,000 g = 42,650 kg

Mass of  magnesium ion in V = $14.8 mg/L\times 3.637\times 10^8 L$

= $5.382\times 10^{9} mg$

=  $5.382\times 10^{6} g$

Moles of magnesium ions = $\frac{5.382\times 10^{6} g}{24 g/mol}=224,250 mol$

From 1 mole of magnesium ion 1 mol of magnesium hydroxide is formed . then from 224,250 moles of magnesium ion will form :

$\frac{1}{1}\times 224,250 mol=224,250 mol$ of magnesium hydroxide

Mass of 224,250 moles of magnesium hydroxide:

224,250 mol × 58 g/mol  = 13,006,500 g = 13,006.5 kg

42,650 kg of calcium carbonate will be produced everyday.

13,600.5 kg of magnesium hydroxide will be produced everyday.