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2 Al(NO3)3 + 3 Na2CO3 = Al2(CO3)3(s) + 6 NaNO3 Use the limiting reagent to determine how many grams of Al2(CO3)3 should precipitate out in the reaction. (2 points)

Chemistry
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so i figured out theat the limiting reagent is Al(NO3)3 since there are 4.667 moles of al(no3)3 and 6 moles of na2co3 needed
Cool. Now use the amount of the limiting reactant to calculate the mass of the precipitate using stoichiometry.

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Other answers:

The precipitate is the solid.
Because, if you're not part of the solution, you're part of the precipitate!! Hahahaha!
:D
I've gotta go now though. I'll be back on later. Have fun!!!

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