Quick question about limiting reagent

This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.

christian15213

Membership Revoked
Removed
10+ Year Member
15+ Year Member
Joined
Jul 28, 2006
Messages
630
Reaction score
1
When figuring a balanced equation with different ratio:s... <<< lol,

Are the ratio's always the indicator of the limiting reagent or is the actual grams per reactants to grams per products what I should be focusing on...

LOL, chem has been a while and everything in the beggining of the EK book has been convienient to think ratio's are the answer but I don't think that is the total trick of the game... I think mols and grams has something to do with it... could anyone elaborate?
 
When figuring a balanced equation with different ratio:s... <<< lol,

Are the ratio's always the indicator of the limiting reagent or is the actual grams per reactants to grams per products what I should be focusing on...

LOL, chem has been a while and everything in the beggining of the EK book has been convienient to think ratio's are the answer but I don't think that is the total trick of the game... I think mols and grams has something to do with it... could anyone elaborate?

there is a little trick which i have no clue why is not used in US (i leaner it in hs in Poland)
You do not need to convert gr to moles and it is acctually a very fast way to calculate what you need. You just use a proportion.
Let's just see it on an example ( i am making up the numbers LOL)
a simple reaction
H2 + O2--> H2O
and let's say they give you 5 grams of H2 and 10 grams of O2
a)and they ask you to find out which one is limmiting reagent?
b)how many moles of H2O will be formed?
c) how manyy grams of O2 or H2 will be left after reaction is complete (of course assuming that the gasses cannot escape into the air from our system)?
SOLUTION:
first thing you need to do is balance the equasion
so you get

2H2 + O2---->2H2O and write what you have given below the reactants
----------------------
5g 10g

using proportion and few calculations:

2 moles of H2= 2x2x1= 4 g
and 1 mole of O2= 16X2=32 g

now proportion using the reaction equasion:

if 4g of H2----reacts with---> 32 g of O2
the how many (xg)-----will react with--->10g
------------------------------------------------------
now cross multiply x=(10x4)/32=1.25g
this means that 1.25 g of H2 will react FULLY with 10 g of O2 which means that there will be H2 left but no more O2 for it to react with so that means that O2 is a limmiting reactant.
It doesn't matter which one you choose to start the problem with either way you get the write answer.

for bart b)
u use the limmining reactant info and again from proportion
you have 10 g to give you -----how many (x) grams of water
if 32g------gives you-------------(2x18)=36
cross multiply
x=(10x36)/32=11.25 g of H2O will be produced.
Now notice that using this method you can limit conversions
Ltes say that they give you the reactants in moles lets say O is still our limmiting and we have 7 moles of it.
now using proportion and stoichiometry of reaction
gives
if 1 mole of O2------>2 moles of H2O
7 moles of O2------>x moles
x=(7x2)/1=14
When using this method always remember about kipping things together ie. O2 under each other and relationships the same across the proportion so if you start with if g---->moles
then you need to follow in the same manner which is g----> moles
i think this is by far the easiest way to calculate things in chemistry. Sometimes using just two proportions you can calculate the whole problem that normally takes a page or so of math where you can make mistakes in every line (and trust me on it i mastered this technique to the point i can do stuff in my head). You can solve any stochiometry related problems using this thing. I mean ANY...and i am not kidding. So this is deffinetly worth knowing.
Hope this helps.
It would be easier to explain this in person since it is a very visual approach but i hope this will be enough. Any questions just ask
:luck:
 
That method is interesting, but it is essentially the same as converting to moles, except it is done in a more indirect sense. Thanks for the input.
 
that seems more complicated than converting to moles.

it does because i had to write out the explanation how to do it but all you really need to solve that problem is 2 proportions.
It's fast and logical and you will never make any conversion error with this method.
 
it does because i had to write out the explanation how to do it but all you really need to solve that problem is 2 proportions.
It's fast and logical and you will never make any conversion error with this method.

yeah...most math is harder to write about than explain, so if it works for you go with it!
 
yeah...most math is harder to write about than explain, so if it works for you go with it!

yeah i love this method because it is almost imposible to make a conversion or math mistake. Unless you can't multiply two numbers and then devide the answear by another number LOL.
It also limits the math since you do not need to convert everything. Let's say you had 5 g of reactant reacting to produce 0.003 moles of product and reaction is 1:1
now you always know molar mass of reactant from PT and let's say it's 10
all you need to do now is
5g------0.003 moles
10g------x moles
x=(10x0.003)/5= 0.006 moles
and you have your answear right away without even having to convert grams to moles first.
For the MCAT this is worth knowing because as you may know converting all that stuff in your head (since we can't use calculators) would be kinda hard and time consuming. And we all know that during MCAT time is everything.
I hope that this can help someone couse it sure does help me be more time efficient and prevents converting errors.
You can also use this method for converting but you don't even have to.
 
Top