Question

How many grams of FeSO4 dissolved per gives 210.5ppm of hardness?
(Fe=56 S=32 O=16 Ca=40 C=12)​

Answer 1

Answer:

How many gm of FeSO4 dissolved per liter gives 210 ppm of hardness?

Two problems :

1) I do not recognise the metric unit gm - this would be gram.metres - but this has little use in chemistry

Does FeSO4 cause hardness in water ? I cannot find any reference to this?

Hardness is generally expressed in terms of Equivalents of CaCO3.

ppm stands for parts per million i.e mg/lit.

Hardness equivalents of CaCO3

= mass of substance causing hardness * equivalent weight of CaCO3/equivalent weight of hardness causing substance.

FeSO4 = Mol.wt = 152 gm , Equivalent weight = 76 gm.

CaCO3 = Mol wt = 100gm , Equivalent weight = 50 gm.

Considering 210 ppm of hardness mentioned is in terms of equivalents of CaCO3.

210 mg/lit= x *50 gm /76 gm

x= 210 mg/lit * 0.6579

x=319.19 mg in one Lit.

i.e 0.319 gm of FeSO4 dissolved in one litre of water has hardness of 210 ppm.

(Hardness expressed in CaCO3 equivalents)

Answer 2

Given:

FeSO4 dissolved per gives 210.5ppm

Fe=56 S=32 O=16 Ca=40 C=12

$\mathrm{FeSO}_{4} \quad \equiv \quad \mathrm{CaCO}_{3}$

$56+32+64=152 \mathrm{mg}\ \ \ \ \ 100g$

$\therefore 100 \mathrm{ppm} \text { of hardness }=152 \mathrm{ppm} \text { of } \mathrm{FeSO}_{4}$

$\therefore 100 \mathrm{ppm} \text { of hardness }=152 \mathrm{ppm} \text { of } \mathrm{FeSO}_{4}$

$210.5 \text { ppm of hardness }=\frac{152 \times 210.5}{100}$

$=319.96 \mathrm{ppm} \text { of } \mathrm{FeSO}_{4}$

$=319.96 \mathrm{mg} / \mathrm{L} \text { or } 0.3199 \mathrm{~g} / \mathrm{L} \text { of } \mathrm{FeSO}_{4}$

Hence the answer is $319.96 \mathrm{mg} / \mathrm{L} \text { or } 0.3199 \mathrm{~g} / \mathrm{L} \text { of } \mathrm{FeSO}_{4}$