November 01, 2024, 04:18:42 AM
Forum Rules: Read This Before Posting


Topic: stoichiometry/limiting reagents  (Read 6587 times)

0 Members and 1 Guest are viewing this topic.

Offline nikita

  • Regular Member
  • ***
  • Posts: 64
  • Mole Snacks: +1/-0
stoichiometry/limiting reagents
« on: September 20, 2008, 09:02:12 PM »
I have my first chem 1 exam on friday and I have been going through all my professors practice exams, so this is not homework.  I also have all the answers, so these are just questions that I seem to keep either getting wrong, or I am simply confused as to what is being asked, rather than not knowing how to do the problem.  I havent taken chemistry in over 10 years, I am an adult back in chemistry, so please be gentle. I am definitely embarrassed about some of my questions.   I hope my title was specific enough, as there are a few different topics, but they are all rather geared toward the same topic.

1.  a 2.000g sample of an oxide of bromine is converted to 2.936g of silver bromide.  What is the empirical formula of the oxide?
*i know the answer is BrO3.  i also know that  2g BrO3 x 1 mol BrO3/ 127.9 g BrO3 = 2.936 g AgBr
however, this is after the fact.  How would I set this up?  Im sure its a question of algebra, but I cannot figure it out.  I know i could figure most of these out by trial and error, but id like to know the way.

2.  Aluminum and bromine react vigorously as represented by the following equation:
           2 Al(s) + 3 Br2(l)  :rarrow: 2 AlBr3(s)
     What mass of product, in grams, can be made by reacting 5.0 g of aluminum and 22 g of bromine?
*i thought this was a stoichiometry prob of limiting reagents.  am i totally wrong?  this is how i am doing it:
5g Al x  1 mol Al/26.98g  x  2 mol AlBr3/2 mol Al  =  .741 mol AlBr3
22g Br2 x 1 mol Br2/159.8 g x 2 mol AlBr3/3 mol Br2 = .826 mol AlBr3

so the limiting reagent is Al
.741 mol AlBr3 x 266.68 g AlBr3/ 1 mol AlBr3 = 197.61 g AlBr3
totally wrong.  the answer is 24 g.  am i totally wrong setting it up?  is there an egregious math error? 

3.  this might be better suited for a math forum, but im here and its about molar mass, so.. (and yes, im in precalc, and i cant figure this out.)
A sample of compound M3O4 weighing 30.0 grams contains 20.0 grams of M. What is
the molar mass of M?
* i have no idea how to set this up. 

ill stop there.  Id appreciate any kicks in the right direction.  thank you.  (ps, i cannot preview my question, so i apologise for any formatting errors.)

Offline Yggdrasil

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 3215
  • Mole Snacks: +485/-21
  • Gender: Male
  • Physical Biochemist
Re: stoichiometry/limiting reagents
« Reply #1 on: September 20, 2008, 11:42:02 PM »
1.  a 2.000g sample of an oxide of bromine is converted to 2.936g of silver bromide.  What is the empirical formula of the oxide?
*i know the answer is BrO3.  i also know that  2g BrO3 x 1 mol BrO3/ 127.9 g BrO3 = 2.936 g AgBr
however, this is after the fact.  How would I set this up?  Im sure its a question of algebra, but I cannot figure it out.  I know i could figure most of these out by trial and error, but id like to know the way.

Because of the conservation of matter, the number of moles of bromine atoms in 2.000g of your bromine oxide will be equal to the moles of bromine atoms in 2.936g of silver bromide.

Quote
2.  Aluminum and bromine react vigorously as represented by the following equation:
           2 Al(s) + 3 Br2(l)  :rarrow: 2 AlBr3(s)
     What mass of product, in grams, can be made by reacting 5.0 g of aluminum and 22 g of bromine?
*i thought this was a stoichiometry prob of limiting reagents.  am i totally wrong?  this is how i am doing it:
5g Al x  1 mol Al/26.98g  x  2 mol AlBr3/2 mol Al  =  .741 mol AlBr3
22g Br2 x 1 mol Br2/159.8 g x 2 mol AlBr3/3 mol Br2 = .826 mol AlBr3

You're setting up the problem correctly (it is a limiting reagent question, good job for recognizing this), but the numbers that you are getting here are incorrect.  Check your math.

Quote
3.  this might be better suited for a math forum, but im here and its about molar mass, so.. (and yes, im in precalc, and i cant figure this out.)
A sample of compound M3O4 weighing 30.0 grams contains 20.0 grams of M. What is
the molar mass of M?
* i have no idea how to set this up. 

Can you figure out how many moles of oxygen will be in your 30.0g sample?

Offline nikita

  • Regular Member
  • ***
  • Posts: 64
  • Mole Snacks: +1/-0
Re: stoichiometry/limiting reagents
« Reply #2 on: October 04, 2008, 04:58:24 PM »
Hi
Thanks for answering.  The reason why I was making so many numerical mistakes was because i was idiotically not using my calculator correctly.  i now put everything in parentheses.  Its an error i never even thought was entering my problems.  duh!

and now, looking back at my other questions, i feel so dumb for asking them, but im glad i know how to do that stuff now.  i got an 88 on my exam and the average was a 52, so i must be doing something right.

thanks for your time!

Sponsored Links