1. Limiting Reactants
Why Reactions Stop

2. p. 8 of Notes Why Do Reactions Stop? One Valentine Treat Bag requires:
1 bag with closure
1 box Sweet Heart candies
2 watermelon Blow-Pops
4 Hershey’s Kisses
3 Hershey’s Hugs
1 heart-shaped Valentine card
Based on the inventory of supplies listed on p. 8, how many treat bags can be assembled?
Why can’t additional bags be put together?
The ____________________ represent the limiting reactant. All other reactants are ____________ reactants. 24
Hershey’s Hugs
excess

3. Terms Defined p. 8 of Notes LIMITING REACTANT (LR) A reactant that:
is totally consumed during a chemical reaction,
limits the extent of the reaction, and
determines the amount of product
EXCESS REACTANT (XR)
A reactant that remains after a chemical reaction stops

4. Why Do Reactions Stop? p. 8 of Notes
Treat bags stopped being assembled because there were no more Hershey’s Hugs in the supply inventory. Clothes stop being purchased because there is no more money in my bank account. A product stops being formed because there is no more limiting reactant available for the reaction.

5. 1 mol MgO + 1 mol H2O → 1 mol Mg(OH)2
p. 8 of Notes
MgO + H2O → Mg(OH)2
Interpret this equation in terms of moles.
1 mol MgO + 1 mol H2O → 1 mol Mg(OH)2
When 1 mole of magnesium oxide is combined with 1 mole of water, 1 mole of magnesium hydroxide will be synthesized.
In other words, to produce 1 mole of magnesium hydroxide, 1 mole of magnesium oxide and 1 mole of water are required.
When the reaction stops, all the magnesium oxide and water has been used up and 1 mole magnesium hydroxide has been formed.

6. MgO + H2O → Mg(OH)2 1 mol H2O 1 mol MgO 1 mol Mg(OH)2 1 mol H2O + →
p. 8 of Notes
MgO + H2O → Mg(OH)2
If 2 moles MgO reacts with 3 moles water, which reactant will be used up first?
MgO
1 mol H2O
1 mol MgO
1 mol Mg(OH)2
1 mol H2O + →
1 mol MgO
1 mol Mg(OH)2
1 mol H2O

7. MgO + H2O → Mg(OH)2 1 mol Mg(OH)2 + → 1 mol Mg(OH)2 1 mol H2O
p. 8 of Notes
MgO + H2O → Mg(OH)2
Which substance is available in excess? How much is available in excess?
H2O
1 mole
1 mol Mg(OH)2 + →
1 mol Mg(OH)2
1 mol H2O

8. MgO + H2O → Mg(OH)2 1 mol Mg(OH)2 + → 1 mol Mg(OH)2 1 mol H2O
p. 8 of Notes
MgO + H2O → Mg(OH)2
The amount of magnesium hydroxide that can be produced depends upon the amount of which reactant?
MgO
1 mol Mg(OH)2 + →
1 mol Mg(OH)2
1 mol H2O

9. Identify Reactants p. 8 of Notes
Balance the equation for the single replacement reaction. ____ KI + ____ Cl2 → ____ KCl + ____I2 Interpret this equation in terms of moles. 2 mol KI + 1 mol Cl2 → 2 mol KCl + 1 mol I2 When given 5 moles of KI and 5 moles Cl2, identify the limiting and excess reactants. LR: XR: 2 KI
Cl2

10. 2 KI + Cl2 → 2 KCl + I2 p. 8 of Notes 5 mol 5 mol ? mol
When starting with 5 moles KI and 5 moles Cl2, how many moles potassium chloride will be produced? The amount of product is determined by the limiting reactant. LR: KI XR: Cl2 Use the limiting reactant as the GIVEN.

11. 2 KI + Cl2 → 2 KCl + I2 p. 8 of Notes 5 mol 5 mol ? mol 5 mol KI
UNKNOWN GIVEN g→mol (G) Mole Ratio mol→g (U)
5 mol KI
2 mol KCl
mol KCl =
2 mol KI
= 5 mol KCl

12. LR Homework
Limiting Reactant HW
p. 380: #77, #78

13. % Yield
Percent Yield
THEORETICAL YIELD maximum amount of product expected from reaction with given amount of reactants ACTUAL YIELD amount of product actually formed during reaction in an experiment

14. % Yield
Percent Yield
DEFINED measure of the efficiency of a reaction calculation of percentage of the amount of product that could theoretically be produced
actual yield
part
% yield =
x 100
theoretical yield
whole

15. Percent Yield % Yield % yield = actual yield theoretical yield x 100
ACTUAL YIELD (AY) → from experimental results THEORETICAL YIELD (TY) → from stoichiometric calculations

16. Percent Yield Practice
Practice/HW
Percent Yield Practice
p. 372: #27

17. Percent Yield Practice
Aluminum hydroxide is often present in antacids to neutralize stomach acid (HCl). Al(OH)3 + 3 HCl → AlCl3 + 3 H2O  If 14.0 g aluminum hydroxide is present in an antacid tablet, determine the theoretical yield of aluminum chloride produced when the tablet reacts with stomach acid.  If the actual yield of aluminum chloride from this tablet is 22.0 g, what is the percent yield?

18. Al(OH)3 + 3 HCl → AlCl3 + 3 H2O Practice #27 14.0 g ? g 1 mol AlCl3
UNKNOWN GIVEN g→mol (G) Mole Ratio mol→g (U)
14.0 g Al(OH)3
1 mol Al(OH)3
1 mol AlCl3
133.5 g AlCl3
g AlCl3 =
78.0 g Al(OH)3
1 mol Al(OH)3
1 mol AlCl3
= 24.0 g AlCl3

19. Percent Yield Practice #27
Aluminum hydroxide is often present in antacids to neutralize stomach acid (HCl). Al(OH)3 + 3 HCl → AlCl3 + 3 H2O  If 14.0 g aluminum hydroxide is present in an antacid tablet, determine the theoretical yield of aluminum chloride produced when the tablet reacts with stomach acid.  If the actual yield of aluminum chloride from this tablet is 22.0 g, what is the percent yield?
24.0 g AlCl3

20. Percent Yield Practice #27 % yield = actual yield theoretical yield
x 100
ACTUAL =
THEORETICAL =
22.0 g AlCl3
24.0 g AlCl3
22.0 g AlCl3
91.7%
AlCl3
% yield =
x 100 =
24.0 g AlCl3

21. Percent Yield Practice/ HW