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anonymous
 one year 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
 one year 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
 one year 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
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.0The precipitate is the solid.

JoannaBlackwelder
 one year 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
 one year 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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