chemistry analyze the amount of softener

Welcome back.

and 2 others joined a min ago.

789views

written 2.9 years ago by

yashatulkar688 • 20

modified 2.9 years ago by

Krishna_Agrawal • 2.7k

(b) Analyze the amount of lime and soda needed for softening 50000 liter of water containing the following: Mg(HCO3)2: 52.00 mg/L, Ca(HCO3)2 : 35.2 mg/L, CaCl2 : 22.2 mg/L, KCl : 15.55 mg/L

engineering chemistry

ADD COMMENT EDIT

1 Answer

59views

written 2.9 years ago by

Krishna_Agrawal • 2.7k

Impurities (mg/lit)	Multiplication factor	CaCO3 equivalent (mg/lit)
$Ca(HCO_3)_2 = 35.2\ mg/lit $	$\frac{100}{162}$	$ 35.2 \times \frac{100}{162} = 21.728$
$Mg(HCO_3)_2 = 52\ mg/lit $	$\frac{100}{146}$	$52 \times \frac{100}{146} = 35.616 $
$CaCl_2 = 22.2\ mg/lit $	$\frac{100}{111}$	$22.2 \times \frac{100}{111} = 20$
$KCl = 15.55\ mg/lit $	-	-

$Lime = \frac{74}{100} [CaCO_3\ equivalent\ of\ Ca(HCO_3)_2 + 2 \times Mg(HCO_3)_2 ] \times \frac{Vol. of water}{1000} \times \frac{100}{\% purity} $

$ = \frac{74}{100} \times [21.728 + 2\times35.616] \times \frac{50,000}{1000} \times \frac{100}{90}$

$ = 3821.688 grams $

$ Soda = \frac{106}{100} \times [20 ] \times \frac{50,000}{1000} \times \frac{100}{95}$

$ = 1115.78 grams $

ADD COMMENT EDIT

Community

Content

Company