I thought I would start out in saying that I think this is a random, but great question. This is because as you said there's so little information out there regarding this sort of reaction so it was an interesting little puzzle for me to figure out for you. My guess would be that the reaction you detailed would not occur. This double replacement reaction would occur in aqueous solution as otherwise the product of magnesium bicarbonate probably wouldn't have a chance of forming since at least one other alkali earth metal bicarbonate salt, calcium bicarbonate, does not exist outside of aqueous solution. I tell you this piece of information instead of whether magnesium bicarbonate exists outside of aqueous solution because I just don't know for certain whether it does. Evidence for this theory that it doesn't exist out of aqueous solution includes (in addition to the point raised above about calcium bicarbonate):
Since it occurs in aqueous the only driving force to the reaction would be the production of an insoluble salt, in your reaction, however, there is no insoluble products that results. Therefore I suspect an alternate reaction would occur. Namely: $$\ce{2NaHCO_3 + MgCl_2 -> 2NaCl + MgO + H_2 O + 2 CO_2}$$ As this reaction would lead to three insoluble products and thus would invoke a greater drive for the reaction. You'd need 0.500 ML in order to precipitate that much magnesium from your solution. Start with the balanced chemical equation for this double replacement reaction #MgCl_(2(aq)) + color(red)(2)NaOH_((aq)) -> Mg(OH)_(2(s)) darr + 2NaCl_text((aq])# This reaction will form sodium chloride, a soluble salt, and magnesium hydroxide, an insoluble solid that will precipitate out of the solution. Notice the #1:color(red)(2)# mole ratio that exists between magnesium chloride and sodium hydroxide. This tells you that, for every 1 mole of magnesium chloride, you need #color(red)(2)# moles of sodium hydroxide in order for all of the magnesium to be removed from the solution. So, you know that you're dealing with a 1.00-ML, 0.0500-mol/L solution of magnesium chloride. Use the solution's molarity to determine how many moles of magnesium chloride you have #C = n/V => n = C * V# #n_(MgCl_2) = 0.0500"mol"/cancel("L") * 1.0 * 10^(-6)cancel("L") = 0.0500 * 10^(-6)"moles"# #MgCl_2# Use the aforementioned mole ratio to determine how many moles of sodium hydroxide you'd need #0.0500 * 10^(-6)cancel("moles"MgCl_2) * (color(red)(2)"moles"NaOH)/(1cancel("mole"MgCl_2)) = 0.100 * 10^(-6)"moles"# #NaOH# Since you know the molarity of the sodium hydroxide solution, you can determine the volume you'd need by #C = n/V => V = n/C# #V_(NaOH) = (0.100 * 10^(-6)"moles")/(0.200cancel("mol")/"L") = 0.5 * 10^(-6)"L"# This is equivalent to having #0.5 * 10^(-6)cancel("L") * "1 ML"/(10^(-6)cancel("L")) = color(green)("0.5 ML")# SIDE NOTE Here's how the net ionic equation looks like for your reaction #Mg_((aq))^(2+) + 2OH_((aq))^(-) -> Mg(OH)_(2(s)) darr#
This is a precipitation reaction: Mg(OH)2 is the formed precipitate. Reactants:
Products:
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