What amount of ferrice chloride is required for the treatment of 1 cubic metre wastewater of   dying industry?
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Ferric chloride dosages can range from 100 to 3000 mg/L depending on the wastewater and other treatments present.
1185923
You will need to hit the library since such general information will require multiple books to explain.  If you have some specific questions, post those.  Just the dyeing part is a very complicated subject if you want details.  If you want some general starting points look at the links at \n\nhttp://www.itt.edu/staff/boyter/links/index.html#WASTEWATER
566520
Fe(s)+H2SO4(aq)excess--> Fe2+(aq) + SO42-(aq)+H2(g)\n\nratio is 1:1\n\n5gFe/55.85g/molFe= .090mol\n\nand\n\n1 molar solution =1 mole per liter\n\nso,\n\n.09mol(1 liter/1mol)= .09L or 90mls
1172374
You could try passing it through carbon, but that may also remove iron.  You would need to try it with known solutions.\nYou may need to go to AA or ICP analysis.
318669
The obvious answer "by destillation" is technically challenging. I know of a Danish firm EnvoTech ApS that built systems which concentrate problematic waste water by cyclical destillation with very little energy loss,  which would help concentrate the remaining acid, but only up to a degree where the stainless steel the system is built from begins to corrode, and without removing the discoloration. You'd get quite good water condensate to reuse in the process, which isn't that helpful for your problem...\n\nThe foamy particles might be removable by letting the water go through a flotation trough. The resulting acid will still be weak and brown, though.
273275
Well you can estimate COD by using K2Cr2O7. Add excess to the sample. Then titrate with FeSO4 to get the end point. Then run a blank titration. The difference in amount will give you the requred quantity of K2Cr2O7 required for the titration giving you the COD. The titation link is given below. \nThe method for COD is given in the second link. \nThanks\n\nHope to get voted the best!!
770958
1/1000 x molar mass (grams/litre) and solute in 1 litre
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Ferrous Sulfate is an iron supplement and Sodium Bicarbonate could be baking soda.\n\nMix these two, and it's a wastwater treatment.\n\nhttp://www.nmfrc.org/bluebook/sec624.htm
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I dont know the exact calculation but this site might be able to help you.\n\nhttp://www.fseric.demon.co.uk/skills/salinity.htm
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Ca(OH)2 \n0.185 g/100cc at 0deg C\n        0.077 g/100cc at 100deg C
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5.6104 us gallons\n4.6716 uk gallons\n\n1 cubic yard is:\n\n201.97 us gal\n168.18 uk gal
707894
Cu 2+ ClO3 1-\n\nCu(ClO3)2
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FeCl3 is the limited reactant and the mass of the excess \nreactant present in the vessel is .55g Mg
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a)4 Fe + 3O2--->2 Fe2O3\n\nCH4 + 2O2 ---> CO2 +2 H2O
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The reductive dissolution of goethite ( -FeOOH) and hematite ( -Fe 2 O 3 ) by phenolic reductants has been examined in order to improve the understanding of iron transformations in soils, sediments and aquifers. Rates of goethite reductive dissolution by hydroquinone increased as the pH was increased from pH 1.8 to 4.65, arid the following reaction stoichiometry was obeyed: 2 - FeOOH + QH 2 = 2 Fe 2+ + Q + 4 OH - . As the pH was increased from pH 4.5 to 6.0, the reductive dissolution rate decreased to below the detection limit. At pH 3.4, the reductive dissolution of hematite was two orders of magnitude slower than goethite. The relationship between structure and reactivity was examined for a series of mono-, di-, and tri-hydroxybenzene reductants. Rates of reductive dissolution decreased in the following order: catechol ~- hydroquinone 〉 3,4-dihydroxybenzoic acid 〉 resorcinol-phenol-4-hydroxybenzoic acid.
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Heating the sand with a strong solution of Hydrochloric acid will remove most of the impurities:\n\nFe2O3 + 6HCl --> 2FeCl3 + 3H2O\n\nAl2O3 + 6HCl --> 2AlCl3 + 3H2O\n\nThe chlorides formed can then be removed by rinsing with warm water.\n\nSilica does not react.
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Ah yes, the classic Iron Thiocyanate complexation lab.\n\nTo determine the concentrations for the first part, you need to know the concentrations of the stock solutions and the final volume of the solution you made.  You can then use the formula:\n\nC1*V1 = C2*V2\n\nWhere C is concentration and V is volume.\n\nFor the second part, I'm not really sure what you did in your lab.  Usually you use the calibration curve you constructed from the first part to measure the concentrations in the second part.  There is alot of ugly calculations to be done in there.  Check your lab packet for more information.