1. Chapter 9 Chemical Quantities

2. 9.6 The Concept of Limiting Reactants
Just as if you were to make sandwiches there will be something left over,
Impossible for every atom/molecule to react, so…
must add one reactant in there in excess (xs) to react as much of the other as possible

3. what is the limiting ingredient here. i. e
what is the limiting ingredient here? i.e. what will stop you from making more sandwiches?

4. the “limiting reactant” here were hammers
how many sets of 1 pliers, 2 screwdrivers, and 1 hammer can you make, and what is left over?

5. Mg reacts w/ O2 in 2:1 ratio
So the 6 Mg on left only need 3 O2’s; but there are 7 O2’s!
what happens to the remaining oxygens???

6. 4 O2’s will not play, but they assure that (essentially) all Mg’s will be found and destroyed!
Mg’s will run out first; they limit the amount of product that can be made; Mg is the limiting reactant

7. Here N2 is in excess, H2 was the limiting reactant

8. 9.7 Calculations Involving a Limiting Reactant
If reactants in a reaction are not present in their mol ratios, one will be used up before the other = limiting reactant; the other one = excess
but you can still predict how much prodruct will be made; yippee!
[hint: limiting reactant probs betray themselves in asking how much product you can get when they give amts of 2 reactants]

9. You must determine which of the 2 is limiting and finish the prob w/ that one only, like this:
1) change amts of both reactants (masses) to mols (old stuff)
2) check mol ratios for which one will be used up first (limiting reactant), (new thing!)
3) use that one only to finish prob (forget about the xs reactant) (old stuff)

10. example We’ll start simple:
1.21 mol zinc are added to 2.65 mol HCl. Zinc chloride and hydrogen gas are formed. Which reactant is in XS? Calculate the amt of mols of zinc chloride produced.
START W/ BALANCED EQUATION!

11. example cont’d Zn + 2HCl  ZnCl2 + H2 the Zn:HCl is 1:2
Only one reactant can take us to product (= LR)
Will it be Zn or HCl???

12. example cont’d Zn + 2HCl  ZnCl2 + H2
One way is to pretend you don’t know about one of the reactants, e.g...
If there are 1.21 mol Zn, how many mols of HCl would you need?
the Zn:HCl is 1:2, so 1.21 mol Zn need twice as much HCl = 2.42 mol HCl

13. example cont’d Zn + 2HCl  ZnCl2 + H2 Do we have 2.42 mol HCl???
YESSIREEBOB!, and more! (2.65 mol)
We have plenty of HCl (= XS) :)
but we will run out of Zn (= LR) :(
Zn will take us to products!...

14. example cont’d Zn + 2HCl  ZnCl2 + H2 then, back to the old ways...
1.21 mol Zn
1 mol Zn
1 mol ZnCl2
= 1.21 mol ZnCl2

15. example Zn + 2HCl  ZnCl2 + H2 What if reactants are given in grams?
No problem! Just change to mols first...
79.1 g Zn react with 76.5 g HCl. 1) Which is Limiting Reactant? 2) How much H2 will be formed?

16. example cont’d Zn + 2HCl  ZnCl2 + H2 Balanced equation!
Change the reactants to mols and compare: 79.1 g Zn  1.21 mol Zn 76.5 g HCl  2.10 mol HCl

17. example cont’d Zn + 2HCl  ZnCl2 + H2
pretend not to know about one of the reactants, this time Zn
if there is a 2 HCl : 1 Zn ratio, and we have 2.10 mol of HCl, we should have at least half as much Zn (1.05 mol) for a complete reaction
we have 1.21 mol Zn = PLENTY! = XS

18. example cont’d HCl will take us to product! = 2.12 g H2 2 mol HCl

19. example
1.00 g Zn reacts with 6.2 x 10-3 mol Pb(NO3)2 to form Zn(NO3)2 and Pb… 1) Which is LR? 2) How much Pb will be formed?
first, the balanced equation!...

20. example cont’d Zn + Pb(NO3)2  Pb + Zn(NO3)2
find the mols of both players:
1.00 g Zn  mol Zn
(done already!)  6.2 x 10-3 mol Pb(NO3)2

21. example cont’d Zn + Pb(NO3)2  Pb + Zn(NO3)2
again, pretend we have only one (Zn)
if we have only mol Zn, we need mol Pb(NO3)2 (1:1 ratio)
we don’t have enough!!! So Pb(NO3)2 is the Limiting Reactant and will take us to the products...

22. example cont’d Zn + Pb(NO3)2  Pb + Zn(NO3)2
same old way to our destination...
.0062 mol Pb(NO3)2
1 mol Pb(NO3)2
1 mol Pb
= 1.3 g Pb
207.2 g Pb

23. what happens when there is not enough of one reactant (l)
not enough O2 means sooty flame

24. 9.8 Percent Yield
Up to this point, all the calculations we’ve done only predicts what we’d get in a perfect world, but…
Reactants aren’t always 100% pure; there are spills, etc., therefore, what we really get will be less

25. The predicted yield is called a theoretical yield (the one we do on paper)
What we actually get is called actual yield (the one we get from the lab)
percent yield is...
percent yield =
actual yield
theoretical yield
• 100

26. example
In the reaction of Zn with HCl, g ZnCl2 are formed, although the theoretical yield was 143 g. What was the % yield?
percent yield = g
143 g
• 100
percent yield = 98.0 %

27. Example For the reaction: K2CO3 + 2HCl  2KCl + H2O + CO2
if 45.8 g K2CO3 are added to xs HCl, 46.3 g of KCl are recovered.
1) What is the theoretical yield?
2) What is the percent yield?

28. example cont’d theoretical yield: = 49.4 g KCl 45.8 g K2CO3
1 mol K2CO3
2 mol KCl
= 49.4 g KCl
1 mol KCl
74.6 g KCl
138.2 g K2CO3

29. example cont’d Now we can find % yield... percent yield = 46.3 g
• 100
percent yield = 93.7 %

30. chemists shoot for 100% but are often satisfied with < 50%
chemists involved in manufacturing, medicine, plastics, etc., constantly strive for the highest profit margin; i.e. they seek the elusive 100% yield