Nice
What can you do with the remaining solution after all the carbon dioxide went out? Because if you can perform other things than it's not that hard figuring out the material.
If it's alkali or alkaline earth, than if you put some solution into a flame it should color it! There are colors for these metals and each one is unique.
MgCl2 however does not color the flame, ZnCl2 and AgCl neither so first of all, you put the substance into HCl solution wait for the carbon dioxide to go away then put a glass stick into the solution and then put the glass stick into a Bunsen-burner's flame.
LiCl - carmine-red color
NaCl - bright yellow-orange color
KCl - violet-purple color
CaCl2 - yellowish-red color
RbCl - yellowish violet color
SrCl2 - scarlet-red color
If the flame does not show any color, you must have MgCl2 or ZnCl2 or AgCl. AgCl is a solid compound should present in the solution as a white thing. MgCl2 and ZnCl2 however is interesting.
Grab some NaOH and start to add it to the solution. Both Zn(OH)2 and Mg(OH)2 is solid should present in the solution like particles, but if you keep adding the NaOH, you should get a clear solution because the Zn makes a hidroxo-complex. Na2[Zn(OH)4] and this complex is soluble in water, so the solution should clear up. With Mg this does not happen so there will be no change. And here you go after these you should have clear evidence what carbonate you have.
Only using algebra you can still figure this out. If you calculate the stoichiometric ratios and put exactly the right amount of carbonate, and right amount of HCl, and then measure the solution and calculate the Xchloride amount, then if it does not add up to the one you calculated in your mind you should start again and do this until you have the right amount of Xchloride solution than you can say what it was.
Let's say you have Li2CO3. The equation is this:
Li2CO3 + 2 HCl -> 2 LiCl + CO2 + H2O
so it says that 1 mole of the carbonate makes a reaction with 2 mole of HCl so you preserve the ratios. You take exactly 18.47 g Li2CO3 (1/4 mole) and take 18.235 g(1/2 mole) of HCl but probably you have a solution of HCl if it's 10 m/m% then you should get 18.235/0.1 (182.35) g of it if it is 20 m/m% then 18.235/0.2 (91.175) etc. and we know that the same amount (1/2 mole) litium-chloride is the product, so we have the water in the HCl, the water from the equation and litium-chloride in the final solution, let's pretend we had 10m/m% HCl. Then we put (182.35-18.235) g water with the HCl into the final solution, the reaction made 1/4 mole because we took 1/4 mole of the substance, 1/4 mole of water is 4.5 g. So we have exactly 168.615 g of water. And the reaction made 1/2 mole of LiCl (again because in the equation you so a 2 multiplier before the LiCl) which is 21.205 g. So we should have exactly 189.82 g solution. If you measure the solution and it is approximately the same, then you had Li2CO3.
It is a pain in the ass task, if you can't do any further tests with the solution after pouring the HCl in it's much easier. If not I suggest you calculate this with every substance and start the lab knowing how much substance you should mix with how much HCl solution and what should be the final solutions weight. You just have to repeat this until you get a match.
I really hope you can do further tests with the solutions because the first method is so much easier.
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