Question

A surface water is coagulated with a dosage of 30 mg/l of ferric chloride and an equivalent dosage of lime. (a) How many kilograms of ferric chloride are needed per liter of water treated? (b) How many kilograms of lime are required assuming that lime is provided as CaO with purity of 70%? (c) How many kilograms of Fe(OH)3 sludge are produced per liter of water treated?

Answer 1

Hi, are you sure that you need values in kilograms? since water is never treated with kilogram quantities of substances. However, since the dosage is 30 mg/L you can simply convert 30 mg to kg and get:

$30 mg \times\frac{1 kg}{10^{6} mg} = 3.0 \times 10^{-5} kg$

first we need the reaction of ferric chloride with lime. remember one mole of lime gives one moles of calcium hydroxide.

$CaO + H_{2}O\rightarrow Ca(OH)_{2}$

$2 FeCl_{3} + 3 Ca(OH)_{2}\rightarrow 2 Fe(OH)_{3} + 3 CaCl_{2}$
since one mole of calcium oxide gives one mole of calcium hydroxide, we have an understanding here that calcium hydroxide is equivalent to calcuim oxide and 2 moles of ferric chloride can reacti with 3 moles of calcium oxide (lime). Now the formula weights are

ferric chloride = 162.2
lime = 56.0

so, 324.4 g (or you can have kg) of ferric chloride react with 168 g of lime

$3.0 \times 10^{-5} kg$ of ferric chloride will react with $3.0 \times 10^{-5} \times \frac{168}{324.4}$ kg of lime, which is $1.55 \times 10^{-5}$ kg of lime at 100 % purity. for a purity of 70 % we will need $1.55 \times 10^{-5} \times \frac{100}{70}$ = $2.21 \times 10^{-5} kg$ of lime.

now for the sludge

$2 FeCl_{3} + 3 Ca(OH)_{2}\rightarrow 2 Fe(OH)_{3} + 3 CaCl_{2}$

formula weight of ferric hydroxide is 106.9

since the mole ratio is one to one.

162.2 g ferric chloride will produce 106.9 g ferric hydroxide

$3.0 \times 10^{-5} kg$ of ferric chloride will produce $3.0 \times 10^{-5} \times \frac{106.9}{162.2}$ = $1.98 \times 10^{-5}$ kg