A mixture of NaCl, SiO2, and CaCO3 is separated following the procedure given in this experiment. Indicate how each of the following procedural changes would affect the amount of the specified substance recovered. Briefly explain.
1. The recovery of CaCo3 was attempted by adding 3M H2So4 instead of 3M HCl to the SiO2/CaCo3 residue.
2. The recovery of CaCO3 was attempted by adding 1MKNO3 solution instead of 1MK2CO3 solution to the filtrate, containing aqueous CaCl2.
1. The recovery of CaCo3 was attempted by adding 3M H2So4 instead of 3M HCl to the SiO2/CaCo3 residue.
2. The recovery of CaCO3 was attempted by adding 1MKNO3 solution instead of 1MK2CO3 solution to the filtrate, containing aqueous CaCl2.
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You don't tell what the procedure is, but I imagine you would first dissolve the NaCl
in distilled water, leaving a residue of insoluble silicon dioxide and CaCO3. Then add
dilute HCl solution to the residue which will react with the CaCO3 to form calcium
chloride, which is water-soluble. Filtration will leave behind the SiO2 and by adding
sodium or potassium carbonate to the CaCl2 solution you can recover the CaCO3.
1. If you use sulfuric acid instead of HCl, it will form calcium sulfate which is not
very soluble in water. (CaCl2 is very water soluble, CaSO4 isn't) So most of the
calcium carbonate will form calcium sulfate and remain behind with the SiO2 and
the amount recovered will be less than it should be.
2. Adding KNO3 solution to the CaCl2 solution will give you a resulting solution
containing K+ ions, Ca++ ions, Cl- ions, and NO3- ions. All possible compounds
from that mixture are very water soluble, so no precipitate will form and you won't
recover any calcium carbonate.
Hope this answers both parts of your question.
in distilled water, leaving a residue of insoluble silicon dioxide and CaCO3. Then add
dilute HCl solution to the residue which will react with the CaCO3 to form calcium
chloride, which is water-soluble. Filtration will leave behind the SiO2 and by adding
sodium or potassium carbonate to the CaCl2 solution you can recover the CaCO3.
1. If you use sulfuric acid instead of HCl, it will form calcium sulfate which is not
very soluble in water. (CaCl2 is very water soluble, CaSO4 isn't) So most of the
calcium carbonate will form calcium sulfate and remain behind with the SiO2 and
the amount recovered will be less than it should be.
2. Adding KNO3 solution to the CaCl2 solution will give you a resulting solution
containing K+ ions, Ca++ ions, Cl- ions, and NO3- ions. All possible compounds
from that mixture are very water soluble, so no precipitate will form and you won't
recover any calcium carbonate.
Hope this answers both parts of your question.