Limiting Reactant Problems Stoichiometry

  • #1
Please help me I am so lost!!

Here's the lesson. My writing is in Blue.

Suppose for your movie you need to make 10 batches of this fake blood. In your freezer you have 9 cans of white grape juice, you have 15 tablespoons of red food color and you have 8 teaspoons of yellow food color. Obviously, you need to go to the store, but how many batches can you make?

*Thinking time*
*
*
*
*
*
Time's up! Got an answer?

Here is the thought process.

9 cans of white grape juice will make 9 batches.

15 tbs. of red food color can make 7.5 batches (2 tbs. each)

8 tsp. of yellow food color can make 8 batches.

So.....The answer must be 7.5 batches if you want to make half a batch.

You just did a "limiting reactant problem!" Aren't you excited?

Woohoo yes I am! Problem is that the questions asked below are not as easy as this.. :frown:


The limiting reactant in the problem above was the red food coloring. When you ran out of that item, you had to stop making blood. It "limited" the amount of product produced.

Problems in chemistry are exactly the same as the blood problem, except you will be dealing with amounts in moles and grams instead of tablespoons and cans.

The problems are almost the same as the stoichiometry problems you did in the last section. The only difference is that each problem will have to be done more than once.

This is as far as it makes sense to me. I'm thinking I need to figure out the ratio of (S) to (O2) but I'm not sure how to do that.

A typical problem might read:

* You have 20.0 g of elemental sulfur, S, and 160.0 grams of O2. What mass of SO2 can be formed?

Notice the similarity to the blood problem and the problem above! You are given amounts of each reactant in this type of problem instead of being told that you have an unlimited amount of one.

The thought process is the same. However in the above case, you need to apply a few chemistry concepts to arrive at the answer.

Study time!

1. Please visit to the following site: Stoichiometry: Limiting Reactant

http://www.chem.tamu.edu/class/majors/tutorialnotefiles/limiting.htm [Broken]

Review the information on Limiting Reactants. Use the information to answer the two problems below that are in red.

1. How many grams SO2 can be formed from 20.0 g of S and 160g O2?

2. How many gramsSO2 can be formed from 20 g S and 15.0 g of O2?


2. Show the balanced equation and your work.

Where should I start? I am that confused; I feel like :cry:

Thanks for your help!!
 
Last edited by a moderator:

Answers and Replies

  • #2
1,444
2
first of all you need to convrt everything in grams to MOLES

secondly
this is the reaction that is going to take place
[tex] S +O_{2} \rightarrow SO_{2} [/tex]

so if u had 1 mol of S and 2 mol of O2, how many moles of SO2 would u form??
use the same concept to your problem, but convert to moles first!
 
  • #3
31
0
Be sure to find the limiting reactant first, before calculating the mass of sulfur dioxide for each respective quantity of reactants.
 
  • #4
4
0
To find the limiting reagent, its somewhat or rather like your blood question, u need to find the number of moles of the substances that are involved.

So, for your Question 1, convert the mass given to the SI units.Remember to always write the correct equation out and BALANCE it or else you will get the wrong mole ratio!!

Equation: S (s) + O2 (g) -> SO2(g)

Now, from the periodic table, you would have found that the MR of the S & O2 to be 32.1 and (16.0 x 2) = 32.0 respectively.
So,
No. of moles in S = mass of S / Mr of S
= 20.0 / 32.1
= 0.6231 mol ( to 4 sig. fig.)

From this value, and usin the ratio that was given in the eqn above, we can deduce that for COMPLETE REACTION of the S will require 0.6231 mole of the O2 as well!!

Now,
No. of moles in O2 = mass of O2/ Mr of O2
= 160 / 32.0
= 5.00 mol
Since from the values, we can see that the O2 is present in excess and hence, the S is the limiting reagent!!

So, since the mole ratio is 1:1:1 respectively in the eqn,
this imply that the no of moles in the SO2 is also 0.6231 mol!!

Findin Mr of the SO2 = 32.1 + 16.0 x 2 = 64.1
Hence, mass of SO2 = 64.1 x 0.6231 = 39.9 g ( to 3 sig. fig.)

Try qns 2 on your own now!!=)
 
  • #5
stunner5000pt said:
first of all you need to convrt everything in grams to MOLES

secondly
this is the reaction that is going to take place
[tex] S +O_{2} \rightarrow SO_{2} [/tex]

so if u had 1 mol of S and 2 mol of O2, how many moles of SO2 would u form??
use the same concept to your problem, but convert to moles first!

So,


S + O2 = SO2
32.066 + 31.9988 = 64.0648
1 mole + 1 mole = 2 moles?

Is that it? Wouldn't it only form one mole of SO2?

Thanks for replying!
 
  • #6
LightMage said:
To find the limiting reagent, its somewhat or rather like your blood question, u need to find the number of moles of the substances that are involved.

So, for your Question 1, convert the mass given to the SI units.Remember to always write the correct equation out and BALANCE it or else you will get the wrong mole ratio!!

Equation: S (s) + O2 (g) -> SO2(g)

Now, from the periodic table, you would have found that the MR of the S & O2 to be 32.1 and (16.0 x 2) = 32.0 respectively.
So,
No. of moles in S = mass of S / Mr of S
= 20.0 / 32.1
= 0.6231 mol ( to 4 sig. fig.)

Ok - I understand up to here. I understand how you're converting to mols.

From this value, and usin the ratio that was given in the eqn above, we can deduce that for COMPLETE REACTION of the S will require 0.6231 mole of the O2 as well!!

But how do we know that the reaction will require only .6231 mols of the 02??

Now,
No. of moles in O2 = mass of O2/ Mr of O2
= 160 / 32.0
= 5.00 mol
Since from the values, we can see that the O2 is present in excess and hence, the S is the limiting reagent!!

So, since the mole ratio is 1:1:1 respectively in the eqn,
this imply that the no of moles in the SO2 is also 0.6231 mol!!

How are you finding the mole ratio to be 1:1:1? And why does it imply that the number of moles in SO2 is .6231 mols? Also, is that possible, considering that S has .6231 mols in the ending reaction? Are there no mols for O2? Sorry. I am terrible at Chemistry.. I just do the best I can but my mind doesn't work this way. I'm great at math but this stuff I cannot wrap my head around.

Findin Mr of the SO2 = 32.1 + 16.0 x 2 = 64.1
Hence, mass of SO2 = 64.1 x 0.6231 = 39.9 g ( to 3 sig. fig.)


Try qns 2 on your own now!!=)

Please help me a little bit more! Thanks so much!!
 
  • #7
2
0
can you guys just post the answer
 
  • #8
2
0
im verry confused. what does all this mean

Now,
No. of moles in O2 = mass of O2/ Mr of O2
= 160 / 32.0
= 5.00 mol
Since from the values, we can see that the O2 is present in excess and hence, the S is the limiting reagent!!

So, since the mole ratio is 1:1:1 respectively in the eqn,
this imply that the no of moles in the SO2 is also 0.6231 mol!!
 
  • #9
Borek
Mentor
28,808
3,309
Be sure to find the limiting reactant first, before calculating the mass of sulfur dioxide for each respective quantity of reactants.

Why? You could as well calculate mass of the product for every reactant - and then select the lowest number.



 
  • #11
14
0
okay i really need help with this...
How many grams SO2 can be formed from 20.0 g of S and 160g O2?
my equation is S + O --> SO2
and i change grams to moles or something like that??
someone pleaseee help me figure this out! im so confused...
 
  • #12
437
1
you have to work in moles. Calculate how many moles of Sulphur and Oxygen you have.

S + O2 ---> SO2

Now, the stoichiometric equation says, 1 mol S reacts with 1 mol O2 to give 1 mol SO2.

say from the number of moles you calculated you got 0.5 mol O2 and 0.75 mol S. (an example)

from this you find that S is in larger amount than O2. Since 1 mol O2 reacts with 1 mol S, the 0.5 mol O2 will completely react with 0.5 mol of S. therefore 0.25 mol of S remains. the limiting reagent is O2.

also, according to the equation, 1 mol of O2 produces 1 mol of SO2.
but you have 0.5 mol of O2 which reacted. this means that 0.5 mol SO2 will be produced.

now you find the mass of the 0.5 mol SO2.
 
  • #13
14
0
ohhhh!!!!
okayy i get it!!!
so for this particular problem the one with the lower number of moles is the limting reagant??
 
  • #14
Borek
Mentor
28,808
3,309
Lower number of moles is not enough - you have to account for stoichiometric coefficients. In this case all stoichiometric coefficients are 1, so just looking at number of moles is enough.

Borek
--
pH meter
 
  • #15
14
0
what are stoichiometric coefficients

and why do i have to account for them as well??
im confused... :confused:
 
  • #16
437
1
say you had H2SO4 + 2NaOH -----> Na2SO4 + 2H2O

you have 0.3 mol H2SO4 and 0.4 mol NaOH.

which one is limiting?
 
  • #17
14
0
the limiting one is H2SO4..right?
 
  • #18
437
1
bah no.....

that's why you have to look at the molar coefficients of each reactant. according to the equation, 2 mol of NaOH will react with 1 mol of H2SO4.

but you have 0.4 mol NaOH. this amount will react with 0.2 mol H2SO4. All the NaOH will be over and 0.1 mol of H2SO4 will remain. hence the NaOH is limiting.

try this one,

2H3PO4 + 3Ba(OH)2 -----> Ba3(PO4)2 + 6H2O

say you have 0.5 mol H3PO4 (phosphoric acid) and 0.9 mol Ba(OH)2 (Barium Hydroxide)

which one is limiting?
 
  • #19
14
0
Ba(OH)2...
because the number infront of H3PO4 is 2...right??
 
  • #20
437
1
you determine it like this:

from equation: 2 mol H3PO4 reacts with 3 mol Ba(OH)2
we have 0.5 mol H3PO4 which will therefore react with 0.75 mol Ba(OH)2.

this shows that Ba(OH)2 is in excess, only 0.75 mol will be used up out of the 0.9 mol present. Phosphoric acid is limiting.

if you had done it the other way round...

3 mol Ba(OH)2 reacts with 2 mol H3PO4.

we have 0.9 mol Ba(OH)2 which would therefore react with 0.6 mol H3PO4.
but we only have 0.5 mol H3PO4. hence H3PO4 is limiting, all the Ba(OH)2 will not react.

understood?
 
  • #21
14
0
ohhhh i get it!!
i was thinking of it like backwards.
thankks so muchh
 
  • #22
14
0
does anyone know how to do enthalpys??
here is the problem i have to do... (its in red)

Assignment Problem
Balance the equation and calculate the enthalpy change for the following reaction:

NH3(g) + O2(g) --> N2(g) + H2O(l)

NH3(g)= -46.11 kj/mol

O2(g) = 0 kJ/mol

N2 = 0 kJ/mol

H20= - 285.830 kJ/mol


i think the balanced equation is 4NH3 + 3O2 --> 2N2 +6H2O, but im not sure how to do the enthalpy change.
 

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