A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

How many grams of aluminu?m will be produced if 12.0g of Ba reacts with 9.0g of Al2(SO4)3

  • This Question is Closed
  1. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    The reaction gives one type of precipitate, i.e. BaSO4 then, # mole of Ba = 12 / 137.3 = 0.0874 mole then # of mole of SO4 2- = 9/342.2 (molar mass of Al2(SO4)3) = 0.0263 mole since mole ratio of Ba: Al2(SO4)3 = 3:1, therefore, Ba is in excess. Thus, only 0.0263 mol Ba reacted to give BaSO4 since 3 mole of BaSO4 will be produced, so # mole of BaSO4 produced = 3 * 0.0263=0.0789 mole gram of BaSO4 = 0.0789 * 233.4 = 18.42g// correct me if I'm wrong ;)

  2. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[3Ba + Al2(SO4)3 \rightarrow 3BaSO4 + 2Al ^{3+}\]

  3. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    This is a problem of limiting reactants

  4. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

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
  • 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.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.