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anonymous
 2 years ago
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)
anonymous
 2 years ago
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)

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anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0so 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

JoannaBlackwelder
 2 years ago
Best ResponseYou've already chosen the best response.0Cool. Now use the amount of the limiting reactant to calculate the mass of the precipitate using stoichiometry.

JoannaBlackwelder
 2 years ago
Best ResponseYou've already chosen the best response.0The precipitate is the solid.

JoannaBlackwelder
 2 years ago
Best ResponseYou've already chosen the best response.0Because, if you're not part of the solution, you're part of the precipitate!! Hahahaha!

JoannaBlackwelder
 2 years ago
Best ResponseYou've already chosen the best response.0I've gotta go now though. I'll be back on later. Have fun!!!
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