Question

Iron can be removed from water by oxidation. The following oxidants are commonly used for iron removal. a. Please calculate: for each mg of Fe removed, what is the mass of oxidant needs? b. And what is the mass of sludge generated for each mg of Fe removed? C. If sludge disposal is a concern, which oxidant would you choose? d. If chemical consumption needs to be minimized, which oxidant would you choose? Oxidant Oxygen Chlorine Chlorine dioxide Potassium permanganate Reaction 4Fe(HCO3)2 + O2 + 2H20 — 4Fe(OH)3 + 8CO2 2Fe(HCO3)2 + Ca(HCO3)2 + Cl2 → 2Fe(OH)3 + CaCl2 + 6CO2 Fe(HCO3)2 + NaHCO3 + CIO2 → Fe(OH)3 + NaClO2 + 3C02 3Fe(HCO3)2 + KMnO4 + 2H20 — 3Fe(OH)3 + MnO2 + KHCO3 + 5C02

Answer 1

All calculations are for 1 mg of iron in the form of Fe(CO₃)₂

a. With oxygen:

1 mg of iron is equivalent to:

$n=\frac{0.001g}{55.85g/mol}=1.79x10^{-5}moles$

For every 4 moles of Fe consumed, one mole of O₂ is consumed. This means that, to consume 1.79x10^-5 moles of Fe, 1/4 of this amount will be needed of oxygen. This is 4.48x10^-6 moles, which are equivalent to a mass of:

$m=n\cdot m_{molar}=4.48x10^{-6}moles\cdot 32\frac{g}{mol}=1.43x10^{-4}g=0.143mg$

b) The number of moles of Fe(OH)₃ produced is the same as the number of moles of Fe that are consumed as Fe(HCO₃)₂, so there will be a generation of 1.79x10^-5 moles of ferric hydroxide, which are equivalent to a mass of:

$m=n\cdot m_{molar}=1.79x10^{-5}moles\cdot 106.85\frac{g}{mol}=0.00191g=1.91mg$

This can be considered as the total mass, since CO₂ is a gas

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For chlorine:

For every 2 moles of Fe consumed, one mole of Cl₂ is consumed. This means that, to consume 1.79x10^-5 moles of Fe, 1/2 of this amount will be needed of chlorine. This is 8.95x10^-6 moles, which are equivalent to a mass of:

$m=n\cdot m_{molar}=9.85x10^{-6}moles\cdot 70.9\frac{g}{mol}=6.35x10^{-4}g=0.635mg$

b) The number of moles of Fe(OH)₃ produced is the same as the number of moles of Fe that are consumed as Fe(HCO₃)₂, so there will be a generation of 1.79x10^-5 moles of ferric hydroxide, which are equivalent to a mass of:

$m=n\cdot m_{molar}=1.79x10^{-5}moles\cdot 106.85\frac{g}{mol}=0.00191g=1.91mg$

And half this amount of moles will be produced of CaCl₂, which is equivalent to a mass of:

$m=n\cdot m_{molar}=8.95x10^{-6}moles\cdot 111\frac{g}{mol}=9.93x10^{-4}g=0.993mg$

The total mass of sludge will be: 290 mg

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For chlorine dioxide:

For every mole of Fe consumed, one mole of ClO₂ is consumed. This means that, to consume 1.79x10^-5 moles of Fe, 1.79x10^-5 moles of ClO₂ are needed, which are equivalent to a mass of:

$m=n\cdot m_{molar}=1.79x10^{-5}moles\cdot 67.45\frac{g}{mol}=1.21x10^{-3}g=1.21mg$

b) The number of moles of Fe(OH)₃ produced is the same as the number of moles of Fe that are consumed as Fe(HCO₃)₂, so there will be a generation of 1.79x10^-5 moles of ferric hydroxide, which are equivalent to a mass of:

$m=n\cdot m_{molar}=1.79x10^{-5}moles\cdot 106.85\frac{g}{mol}=0.00191g=1.91mg$

And this same amount of moles will be produced of NaClO₂, which is equivalent to a mass of:

$m=n\cdot m_{molar}=1.79x10^{-5}moles\cdot 90.44\frac{g}{mol}=1.61x10^{-4}g=1.61mg$

The total mass of sludge will be: 353 mg

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For potassium permanganate:

For every 3 moles of Fe consumed, one mole of KMnO₄ is consumed. This means that, to consume 1.79x10^-5 moles of Fe, 1/3 of that amount is the needed number of moles of permanganate. This means that 5.97x10^-6 moles of KMnO₄ are needed, which are equivalent to a mass of:

$m=n\cdot m_{molar}=5.97x10^{-6}moles\cdot 158\frac{g}{mol}=9.43x10^{-4}g=0.943mg$

b) The number of moles of Fe(OH)₃ produced is the same as the number of moles of Fe that are consumed as Fe(HCO₃)₂, so there will be a generation of 1.79x10^-5 moles of ferric hydroxide, which are equivalent to a mass of:

$m=n\cdot m_{molar}=1.79x10^{-5}moles\cdot 106.85\frac{g}{mol}=0.00191g=1.91mg$

1/3 of amount of moles will be produced of MnO₂, which is equivalent to a mass of:

$m=n\cdot m_{molar}=5.98x10^{-6}moles\cdot 86.93\frac{g}{mol}=5.19x10^{-4}g=0.519mg$

And the same amount of moles will be produced of KHCO₃, equivalent to a mass of:

$m=n\cdot m_{molar}=5.98x10^{-6}moles\cdot 100.1\frac{g}{mol}=5.97x10^{-4}g=0.597mg$

The total mass of sludge will be: 303 mg

In terms of sludge disposals, the method with oxygen is the best, since it results in the least mass of sludge.

In terms of chemical consumption, O₂ is the best as well, since it is the one that requires the less mass. Plus, oxygen is widely available in air for the treatment.