Question

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)

Answer 1

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

Answer 2

@JoannaBlackwelder

Answer 3

Cool. Now use the amount of the limiting reactant to calculate the mass of the precipitate using stoichiometry.

Answer 4

The precipitate is the solid.

Answer 5

Because, if you're not part of the solution, you're part of the precipitate!! Hahahaha!

Answer 6

:D

Answer 7

I've gotta go now though. I'll be back on later. Have fun!!!