Table 1311

Summary of Equivalent Masses

Eqs. (13.74) through (13.78) — 27.45 — — 44.45 43.45

The solids produced from the previous reactions are Fe(OH)2(s), Fe(OH)3W, Mn(0H)2(s), and MnO2(s). From Equation (13.69), the equivalent mass of Fe(OH)2(s) is Fe(OH)2/2 = 44.9 and that of Fe2+ is Fe/2 = 27.9. From Eqs. (13.70) through (13.73), the equivalent mass of Fe(OH)3(s) is Fe(OH)3(s)/1 = 106.8; the corresponding equivalent mass of Fe2+ is Fe/1 = 55.8. Equations (13.74) through (13.77) produce the following equivalent masses: Mn2+ = Mn/2 = 27.45; Mn(OH)2(s) = Mn(OH)2(s)/2 = 44.45 and MnO2(s) = MnO2(s)/2 = 43.45. Table 13.11 summarizes the equivalent masses.

Let MFe(OH) FeRem and MFe(OH^FeRem be the kilogram mass of ferrous hydroxide and ferric hydroxide produced from A[Fe]mg milligrams per liter of ferrous iron removed, respectively. MFe(OH^FeRem is produced at the high pH range and MFe(OH)3FeRem is produced at the low pH range. Also, let Mm(0H)2MnRem and M Mn^MnRem be the kilogram mass of manganous hydroxide and manganic oxide produced from A[Mn]mg milligrams per liter of manganese removed, respectively. Also, MMn(0H)2MnRem is produced at the high pH range and MMlio2unRem is produced at the low pH range. Let the volume of water treated be V cubic meters. Using these information and the equivalent masses derived above produce the following:

MFe(0H)2FeRem = 1(XX)(27 9) = 0.0016A[Fe]mgV (13.78)

M Fe( 0H)3FeRem = ^()16()8(AA([!F5;^8a)gVV = 0.0019A[ Fe ]mg V (13.79)

44 45Ar Mn 1 V

Mn(OH)2MnRem 1000(27 45 )

Mn2+

Example 13.7 A raw water contains 2.5 mg/L Mn. Calculate the solids produced per cubic meter of water treated. The effluent is to contain 0.05 mg/L of manganese and the removal is to be done at the high pH range.

Solution:

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