Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
math92130
Group Title
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)
 2 months ago
 2 months ago
math92130 Group Title
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)
 2 months ago
 2 months ago

This Question is Closed

math92130 Group TitleBest ResponseYou've already chosen the best response.0
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
 2 months ago

math92130 Group TitleBest ResponseYou've already chosen the best response.0
@JoannaBlackwelder
 2 months ago

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

JoannaBlackwelder Group TitleBest ResponseYou've already chosen the best response.0
The precipitate is the solid.
 2 months ago

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

JoannaBlackwelder Group TitleBest ResponseYou've already chosen the best response.0
:D
 2 months ago

JoannaBlackwelder Group TitleBest ResponseYou've already chosen the best response.0
I've gotta go now though. I'll be back on later. Have fun!!!
 2 months ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.