1. Chapter 5 Chemical Reactions

2. The Mole

3. 1 pair of kings = 2 kings
1 dozen eggs = 12 eggs
2 pairs of kings = 2 x 2 kings = 4 kings
2 dozen eggs = 24 eggs

4. The Mole Abbreviation: “mol” 1 mole = 6.022 x 1023 things
Avogadro’s number
Abbreviation: “mol”
The number of carbon atoms in 12g of C-12
How much does this mole of carbon weigh?

5. 1 mole C atoms = 6.022 x 1023 Carbon atoms
6.022 x 1023 Carbon atoms = 1 mole of carbon atoms
1 mole bicycles = x 1023 bicycles
6.022 x 1023 bicycles = 1 mole of bicycles
1 mole of anything = x 1023 items of anything
6.022 x 1023 items of anything = 1 mole of anything

6. Molar Mass The mass, in grams, of one mole of any element or compound
Abbreviated with capital, italicized M
Unit = grams/mole = g/mol
Gives us a way to go from grams to moles

7. Molar Masses for elements and molecules
Use periodic table to figure out Molar mass for each element
Carbon
1 atom of carbon is amu
1mole of carbon is grams
Molar Mass = g/mol
H2O
Made of 2 hydrogen atoms and one oxygen atom
Molar mass of each hydrogen is g/mol
Molar mass of each oxygen is g/mol

8. Problems
1) You need mol Cu for an experiment. How many grams do you need to weigh out? How many atoms of Cu is this?
2) You have g C2H6O. How many moles of C2H6O are present in this sample? How many molecules are present?

9. Chemical Equations H2 (g) + Cl2(g)  2 HCl (g) Reactants Product(s)
Coefficient
Physical State
Subscript
H2 (g) + Cl2(g)  2 HCl (g)
Reactants
Product(s)

10. Law of Conservation of Matter/Mass
Matter is neither created, nor destroyed, but is merely rearranged
The mass of the reactants must equal the mass of the products
Atoms in the reactants must equal atoms in the products
H2 (g) + Cl2(g)  2 HCl (g)

11. Balancing Chemical Equations
Zn(s) + HCl(aq)  H2(g) + ZnCl2(aq)
Write the unbalanced equation
Balance the atoms of one element by placing coefficients in front of molecules (saving single elements for last)
Choose another element and balance it
Continue until all elements have the same number of atoms on both sides of the equation
Check yourself

12. Problems __ N2(g) + __ H2(g)  __ NH3(g)
__ Fe(s) + __ Cl2(g)  __ FeCl3(s)
__NH3(g) + __O2(g)  __NO(g) + __H2O(g)
__C5H12(l) + __O2(g)  __CO2(g) + __H2O(g)

13. General Reactions
Combination rxns: 2 or more substances react to form a single product
2 H2 + O2  2 H2O

14. Decomposition rxns: single substance decomposes into 2 or more products
opposite of combination rxns
2 H2O  2 H2 + O2

16. Single replacement/displacement: one element reacts with a compound to form a new compound and release a new element
2 Na + 2 H2O  2 NaOH + H2

17. Exchange or Double replacement/displacement: an interchange of partners between two compounds
Pb(NO3)2(aq) + K2CrO4(aq)  PbCrO4(s) + 2 KNO3(aq)

18. Combustion rxns: the burning of a compound, usually a hydrocarbon or oxygen containing hydrocarbon, in oxygen to form heat, carbon dioxide and water

19. Problems Identify the reaction type then balance the reaction
__C3H8(g) + __O2(g)  __CO2(g) + __H2O(g)
__BaCl2(aq) + __Na2SO4(aq)  __BaSO4(s) + __NaCl(aq)
__Fe(s) + __H2O(l)  __Fe3O4(s) + __H2(g)
__Pt(s) + __F2(g)  __PtF4(l)
__H3BO3(s)  __B2O3(s) + __H2O(l)
__C4H10(g) + __O2(g)  __CO2(g) + __H2O(g)

20. Stoichiometry
Calculations of the relative quantities of reactants and products in chemical reactions
How much of something do you need to make something else
You learned how to use moles
You learned about chemical reactions and balancing equations.
Now it all comes together!!!

21. Stoichiometry guidelines
Make sure you have a balanced chemical equation
Use the balanced equations to make molar ratios between reactants and products or reactants and reactants or products and products.
Use moles of something to get to moles of something else

22. Molar ratios
Making a conversion factor from a relationship in the chemical equation
Pt(s) + 2F2(g)  PtF4(l)

23. Stoichiometry Flow Chart
Molar ratios: Using coefficients of balanced equation
Moles of A
Moles of B
M = g/mol
M = g/mol
grams of A
grams of B
D = g/mL
D = g/mL
Volume of B
Volume of A

24. Stoichiometry Problems
If you have 10.0g C4H10 (butane), how many grams of water can you make upon combustion?
How much O2 do you need in problem 1 if you’d like to produce 7.39g CO2?
How much CO2 is produced upon combustion of 4.3g of propane (C3H8)?

25. The Jim Bob dilemma
Jim Bob got a job at a chair factory. His boss tells him that he’s going to lunch and when he comes back, Jim Bob better have 5 chairs made or he’s fired. JB gets to work. He counts 32 arms, 18 legs, 4 backs, and 2 seats. Will JB get fired?

26. Limiting Reagent
Limiting Reagent/Reactant/Factor: the reactant/factor that determines the amount of product formed
Other reactants are “in excess”
Cheaper reactants are usually in excess

27. Problems
You have 10.0 moles H2 and 1.00 mol O2. How much H2O can you make?
You combust 10.2 mol propane (C3H8) in 7.80 mol O2. How much CO2 can you produce?

28. If 2. 3 mol carbon disulfide reacts with 5
If 2.3 mol carbon disulfide reacts with 5.4 mol oxygen to form carbon dioxide and sulfur dioxide, what mass of sulfur dioxide is formed?
5.50 g silicon dioxide reacts with 4.71g Carbon to from silicon carbide and carbon monoxide. What mass of carbon monoxide is formed?

29. Jim Bob got a job at a chair factory
Jim Bob got a job at a chair factory. His boss tells him that he’s going to lunch and when he comes back, Jim Bob better have 5 chairs made or he’s fired. Jim Bob gets to work. He counts 32 arms, 18 legs, 4 backs, and 2 seats. Will Jim Bob get fired?
Jim Bob 

30. 2 arms + 4 legs + 1 back + 1 seat  1 chair
32 arms, 18 legs, 4 backs, 2 seats
32arms (1chair/2arms) = 16 chairs
18 legs (1chair/4legs) = 4.5 chairs
4 backs (1chair/1back) = 4 chairs
2seats (1chair/1seat) = 2 chairs

32. Percent Yield
Theoretical Yield: the maximum possible quantity of product
100% yield
Actual/Experimental Yield: quantity of product actually obtained
Percent Yield: efficiency of reaction
PY = (actual yield/theoretical yield) x 100

33. Problems
From the previous question, the theoretical yield of CO is 5.13 g. If you obtained 4.32g CO, what was your % yield?
You react 4.41 mol carbon monoxide with 8.39 mol hydrogen gas to get 122g methanol (CH4O). What is your percent yield?

34. Chapter 5 checklist
Know what the mole is and how to use it as a conversion factor.
Know what molar mass is and how to use it as a conversion factor
Chemical equations – know how to identify components and how to balance an equation
Combination, decomposition, single and double replacement, and combustion reactions.
Stoichiometry- Using the chemical equation to calculate relative amounts of reactants and products
Limiting reagent- you can only obtain product based on which compound you have the least of
Percent yield – How much product did you get compared to what you should of got