1. Symbolic notation used by chemists to represent a chemical reaction.
Chemical Equation
Symbolic notation used by chemists to represent a chemical reaction.

2. In a chemical equation, everything to the left of the arrow
Reactant – starting materials
In a chemical equation, everything to the left of the arrow

3. In a chemical equation, everything to the right of the arrow
Product – ending materials
In a chemical equation, everything to the right of the arrow

4. Reactants  Products Arrow is read as “yields” or “produces”
Equation format

5. Number in front of a formula. Tells how many formula units are present.
Coefficient

6. Balanced chemical equation
Demonstrates conservation of mass. Same # of each element on both sides.
Balanced chemical equation

7. Chemical equations demonstrate conservation of
mass, charge (& energy)
Chemical equations demonstrate conservation of

8. 1. Initial survey: count atoms of each element on each side.
2. Adjust numbers of atoms in each formula using coefficients.
3. Update atom counts after every change.
How to balance equation?

9. Begin with the most complicated formula in the equation & work down to the simplest.
Even numbers are easier to balance than odd numbers.
If a polyatomic ion appears on both sides of the equation, treat it as a unit.
Tricks for balancing

10. If you can’t balance an equation ….
Check the formulas – odds are, at least one formula is incorrect!
If you can’t balance an equation ….

11. Types of Reaction (5) Synthesis Decomposition Single Replacement
Double Replacement
Combustion
Types of Reaction (5)
Regents Level

12. Synthesis
Reaction with 1 product
Regents Level

13. Reaction with 1 reactant
Decomposition
Reaction with 1 reactant
Regents Level

14. Single Replacement
Reaction of 1 element with 1 compound to produce a new element & a new compound
Regents Level

15. Reaction of 2 compounds to produce 2 new compounds.
Double Replacement
Reaction of 2 compounds to produce 2 new compounds.
Regents Level

16. Reaction with oxygen. One of the reactants must be O2.
Combustion
Reaction with oxygen. One of the reactants must be O2.
Regents Level

17. Decomposition
AB  A + B
A & B may be elements, compounds, or one of each
Regents level

18. Synthesis A + B  AB A & B may be elements, compounds, or one of each
Regents Level

19. Single Replacement, Cationic
AX + B  A + BX
A & B, metals or H
Regents level

20. Single Replacement, Anionic
X + AY  Y + AX
X, Y = halogens
Regents level

21. Double Replacement
AX + BY  AY + BX
Regents level

22. Combustion
A + O2  Products
Regents level

23. To predict if a single replacement reaction will occur,
You must compare the reactivity of the free element with the corresponding element in the compound.
Use Table J. If the stand-alone element is above the corresponding element in the compound, the reaction will take place.
To predict if a single replacement reaction will occur,

24. 2Na + MgCl2  2NaCl + Mg Na + MgCl2  ?
Compare Na & Mg (both metals). Na is above Mg in Table J so the rxn occurs.
2Na + MgCl2  2NaCl + Mg
Na + MgCl2  ?

25. F2 + MgCl2  MgF2 + Cl2 F2 + MgCl2  ?
Compare F2 & Cl2 (both nonmetals). F2 is above Cl2 in Table J so the rxn occurs.
F2 + MgCl2  MgF2 + Cl2
F2 + MgCl2  ?

26. Do the following reactions occur?
Br2 + HF: Compare Br2 & F2
Mg + ZnCl2: Compare Mg & Zn
Na + HCl: Compare Na & H2
Ag + LiBr: Compare Ag & Li No
Yes
Yes No

27. 3 kinds of double replacement reactions
1) Those driven to completion by the formation of a precipitate
Those driven to completion by the formation of a gas
Those driven to completion by the formation of a molecular species, often H2O
3 kinds of double replacement reactions

28. 3 kinds of double replacement reactions
Two ionic compounds producing a solid precipitate.
Two compounds producing a gas. Often two ionic compounds or an ionic compound with an acid or base.
Two compounds producing a molecular species. For the 1st year: Acid + Base  water + …
3 kinds of double replacement reactions

29. To predict if a double replacement reaction will occur,
You must determine the possible products & determine if the reaction goes to completion.
The reaction goes to completion if a gas, a solid, or a small covalent molecule (H2O) is formed.
Use Table F to determine if a precipitate (solid) is formed.
To predict if a double replacement reaction will occur,

30. Write formulas for the possible products.
Use Table F to determine if a precipitate (solid) is formed.
Scan through table F looking for information about the solubility of the products.
To predict if a double replacement reaction will produce a precipitate …

31. Ba(NO3)2(aq) + Fe2(SO4)3(aq)  ?
Write formulas for the possible products: Fe(NO3)3 + BaSO4
Scan through table F looking for information about the solubility of the products:
All nitrates are soluble so Fe(NO3)3 is aqueous.
Most sulfates are soluble except when combined with Ag+, Ca2+, Sr2+, Ba2+, & Pb2+ so BaSO4 is solid.
Ba(NO3)2(aq) + Fe2(SO4)3(aq)  ?

32. Stoichiometry Problems
Use the relationship in the balanced equation to predict amounts (moles) consumed or produced.
Use the equation coefficients to set up mole ratios.
Stoichiometry Problems

33. General Procedure for Stoichiometry Problems
1st species into moles
Mole conversion, species 1 to species 2, using coefficients from BCE
2nd species out of moles
General Procedure for Stoichiometry Problems

34. Stoichiometry Problems
May involve mole-mole relationships.
For reactions involving all gases, coefficients in equation also give information about volume-volume ratios.
Stoichiometry Problems

35. Factor-label method
4Na H2 = 2 moles of H2
2Na
2Na + 2H2O  2NaOH + H2 How many moles of hydrogen are produced when 4 moles of sodium reacts completely?
Mole-mole problem

36. Proportion method
The amount of Na is doubled, from 2 moles to 4 moles, so every other term in the equation is doubled.
2Na + 2H2O  2NaOH + H2 How many moles of hydrogen are produced when 4 moles of sodium reacts completely?
Mole-mole
problem

37. Factor-label method 3 L CH4 1 L CO2 = 3 L CO2
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g) How many liters of CO2 are produced when 3 liters of CH4 reacts completely?
Volume-volume problem

38. Proportion Method
The amount of CH4 is tripled, from 1 L to 3 L, so everything else is tripled.
3 liters of CO2 are produced!
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g) How many liters of CO2 are produced when 3 liters of CH4 react completely?

39. Limiting Reactant Problems
Give quantitative info about TWO reactants! Must find LR before doing stoichiometry!
Limiting Reactant Problems

40. NaCl(aq) + AgNO3(aq)  AgCl(s) + NaNO3(aq)
This is a limiting reactant problem because quantitative information about two reactants is given.
A solution with 5.8 g of NaCl is mixed with a solution with 7.2 g of AgNO3.
NaCl(aq) + AgNO3(aq)  AgCl(s) + NaNO3(aq)

41. Limiting Reactant Problems
Requirement: Must have a balanced chemical equation to begin a stoichiometry problem.
Limiting Reactant Problems

42. Limiting Reactant Problems
Two methods to determine LR
1) Tried and true
2) Grouping method
Limiting Reactant Problems

43. 90.0 g < 108. g so H2 = LR!
Tried and True
10.0 g H2 1 mol H2 2 mol H2O 18 g H2O =
2 g H mol H2 1 mol H2O
Calculation #1, based on H2
90. g
Calculation #2, based on O2
108 g
96.0 g O2 1 mol O2 2 mol H2O 18 g H2O =
32 g O mol O mol H2O
How many grams of water can be produced from 10.0 g H2 & 96.0 g O2?
2H2 + O2  2H2O

44. Grouping Method then stoichiometry.
Find moles then groups. Smaller group is LR.
# of groups = # of moles / coefficient
2.5 groups < 3 groups so H2 = LR!
2.5 groups
3 groups
2H2 + O2  2H2O
5 moles
3 moles
10.0 g
96.0 g
How many grams of water can be produced from 10.0 g H2 & 96.0 g O2?

45. How many grams of water can be produced from 10.0 g H2 & 96.0 g O2?
Make sure to use moles of LR, not groups, for given!
For stoichiometry, use MOLES
5 mol H2 2 mol H2O 18 g H2O = 90 g H2O
2 mol H mol H2O
H2 is LR – see previous slide!
2H2 + O2  2H2O
How many grams of water can be produced from 10.0 g H2 & 96.0 g O2?

46. General Procedure for LR problems
Into moles
Identify LR
Mole conversion, using coefficients from BCE. LR is “given” in mole conversion.
Out of moles
Excess reactant, two-step
Calc amt of excess reactant consumed
Subtract amt consumed from initial quantity
General Procedure for LR problems

47. Excess reactant calculation, two-step!
Use stoich conversion to find amt of O2 consumed in reaction
Subtract amt consumed from initial amt
Consider the previous problem:
5.00 mol mol
2H2 + O2  2H2O
10.0 g g
Excess reactant calculation, two-step!

48. Excess reactant calculation, two-step!
Given = LR
Amount consumed
5 mol H2 1 mol O g O2 = 90 g O2
2 mol H2 1 mol O2
Subtract amt consumed from initial amt
96 g -90 g = 6 g O2 in excess.
Consider the previous problem:
5.00 mol mol
2H2 + O2  2H2O
10.0 g g
Excess reactant calculation, two-step!