1. WHERE DO YOU LOOK IF YOU’VE LOST YOUR MIND?
-Bernard Malamud-

2. REVIEW
WHAT INFORMATION DOES THE FOLLOWING FORMULA TELL YOU?
2 (NH4)3PO4

3. WHAT INFORMATION DOES THE FOLLOWING FORMULA TELL YOU?
2 (NH4)3PO4
You have 2 formula units of ammonium phosphate, or
You have 2 moles of ammonium phosphate
In those 2 formula units, you have 6 atoms of nitrogen, 24 atoms of hydrogen, 2 atoms of phosphorus, 8 atoms of oxygen, or
In those 2 moles, you have 6 moles of nitrogen, 24 moles of hydrogen, 2 moles of phosphorus, and 8 moles of oxygen

4. CALCULATE THE MOLAR MASS OF ZnSO4.

5. CALCULATE THE MOLAR MASS OF ZnSO4.
1 Zn = 1 x 65 = 65
1 S = 1 x 32 = 32
4 O = 4 x 16 = 64
Molar Mass = 161 g/mole

6. HOW MANY MOLES OF K2SO4 ARE IN 20 GRAMS?

7. HOW MANY MOLES OF K2SO4 ARE IN 20 GRAMS?
First, calculate the molar mass:
K = 2 x 39 = 78
S = 1 x 32 = 32
O = 4 x 16 = 64
Molar mass = 174 g/mole
# moles = 20 g / 174 g/mole = mole

8. HOW MANY MOLECULES OF C2H5OH ARE IN 2.5 MOLES?

9. HOW MANY MOLECULES OF C2H5OH ARE IN 2.5 MOLES?
1 mole = x 1023 molecules
# molecules = 2.5 moles x x 1023 molecules/mole
= 1.51 x 1024 molecules

10. HOW MANY GRAMS OF PbI2 ARE IN 8.4 X 1022 MOLECULES?

11. HOW MANY GRAMS OF PbI2 ARE IN 8.4 X 1022 MOLECULES?
# moles = 8.4 x 1022 / x 1023
= 1.39 x 10-1 = moles
Molar mass of PbI2 = (127) = 461 g/mole
# grams = 461 g/mole x moles
= 64 g

12. Pb(NO3)2 + AlCl3  Al(NO3)3 + PbCl2 KClO3  KCl + O2
BALANCE THE FOLLOWING EQUATIONS, AND TELL WHAT KIND OF REACTION THE EQUATION REPRESENTS:
C4H10 + O2 → CO2 + H2O
Fe + Cl2  FeCl3
Cu + AgNO3  Cu(NO3)2 + Ag
Pb(NO3)2 + AlCl3  Al(NO3)3 + PbCl2
KClO3  KCl + O2

13. 3 Pb(NO3)2 + 2 AlCl3  2 Al(NO3)3 + 3 PbCl2 2 KClO3  2 KCl + 3 O2
BALANCE THE FOLLOWING EQUATIONS, AND TELL WHAT KIND OF REACTION THE EQUATION REPRESENTS:
2 C4H O2 → 8 CO H2O
2 Fe + 3 Cl2  2 FeCl3
Cu AgNO3  Cu(NO3) Ag
3 Pb(NO3) AlCl3  2 Al(NO3)3 + 3 PbCl2
2 KClO3  2 KCl + 3 O2

14. WHAT ARE THE SUBSTANCES ON THE LEFT SIDE OF A CHEMICAL REACTION CALLED?
WHAT ARE THE SUBSTANCES ON THE RIGHT SIDE OF CHEMICAL REACTION CALLED?

15. WHAT ARE THE SUBSTANCES ON THE LEFT SIDE OF A CHEMICAL REACTION CALLED?
REACTANTS
WHAT ARE THE SUBSTANCES ON THE RIGHT SIDE OF CHEMICAL REACTION CALLED?
PRODUCTS

16. WHAT IS THE LAW OF CONSERVATION OF MASS?

17. WHAT IS THE LAW OF CONSERVATION OF MASS?
IN A CHEMICAL REACTION, MATTER IS NEITHER CREATED OR DESTROYED, ONLY REARRANGED.

18. HOW DOES THE LAW OF CONSERVATION OF MASS APPLY TO A CHEMICAL EQUATION?

19. HOW DOES THE LAW OF CONSERVATION OF MASS APPLY TO A CHEMICAL EQUATION?
YOU HAVE THE SAME NUMBER OF EACH TYPE OF ATOMS ON BOTH SIDES OF THE EQUATION – THE EQUATION IS BALANCED.

20. 2 NH3 + H2SO4  (NH4)2SO4
How many grams of ammonia (NH3) would be required to make 1 kilogram of ammonium sulfate?

21. How many grams of ammonia (NH3) would be required to make 1 kilogram of ammonium sulfate?
X g
2 NH3 + H2SO4  (NH4)2SO4
X / 34 = / 128
X = (1000 x 34) / 128 = g
Note: molar mass of ammonium sulfate = 128 g/mole
molar mass of ammonia = 17 g/mole

23. How many moles of iron, Fe, would be required to make 40 grams of hydrogen, H2?
2 Fe + 6 HCl  2 FeCl H2

24. How many moles of iron, Fe, would be required to make 40 grams of hydrogen, H2?
2 Fe + 6 HCl  2 FeCl H2
Molar mass of H2 = 2 g/mole
# moles H2 = 40 g/2 g/mole = 20 moles X
X/2 = 20/3 so X = 20/3 x 2 = 13.3 moles

25. A COMPOUND CONTAINS THE ELEMENTS COPPER AND CHLORINE IN A RATIO OF 1 COPPER : 2 CHLORINE.
WHAT IS THE EMPIRICAL FORMULA FOR THIS COMPOUND?
WHAT IS THE % COMPOSITIONOF COPPER?
WHAT IS THE % COMPOSITION OF CHLORINE?

26. EMPIRICAL FORMULA: CuCl2
Molar mass = (35.5) = g/mole
% Cu = (63.5 / 134.5) x 100 = 47.2 %
% Cl = (71 / 134.5) x 100 = 52.8 %

27. IN A CHEMICAL REACTION, THE REACTANT IN THE LEAST SUPPLY WILL LIMIT THE AMOUNT OF PRODUCT FORMED.
WE CALL THE REACTANT IN THE SMALLEST SUPPLY THE LIMITING REAGENT.
THE REACTANT IN THE GREATEST SUPPLY IS THE EXCESS REAGENT.

28. 2 Cu + S  Cu2S
WHAT IS THE LIMITING REAGENT WHEN 80.0 G Cu REACTS WITH 25.0 G S?
# moles Cu = 80 g/63.5 g/mole = 1.26 moles
# moles S = 25.0 g/32 g/mole = moles
1.26 X
2 Cu + S  Cu2S /2 = X/1 x = 0.63
So, it would take 0.63 moles of S to react with 1.26 moles Cu. You have moles S, so Cu is the limiting reagent and S is in excess.

29. ANOTHER CALCULATION OF INTEREST IN REACTIONS IS % YIELD.
THE % YIELD IS A MEASURE OF THE EFFICIENCY OF THE REACTION PROCESS.
IF YOU WERE CARRYING OUT THE REACTION ON AN INDUSTRIAL OR COMMERCIAL SCALE, THE % YIELD WOULD BE IMPORTANT.
THINGS CAN HAPPEN IN BOTH THE LABORATORY AND IN PRODUCTION SO THAT REACTIONS DON’T NECESSARILY GO TO COMPLETION.

30. THE % YIELD IS
= (ACTUAL AMT/THEORETICAL YIELD) X 100
THE ACTUAL YIELD IS EXPERIMENTALLY DETERMINED.
THE THEORETICAL YIELD IS CALCULATED FROM THE REACTION EQUATION.

31. YOU HEAT CALCIUM CARBONATE, CaCO3, TO PRODUCE LIME, CaO.
CaCO3  CaO + CO2
IF YOU START WITH 24.8 G OF CaCO3 , AND YOU PRODUCE 10.6 g OF CaO, WHAT IS THE % YIELD.
FIRST, YOU NEED TO CALCULATE THE THEORETICAL YIELD.
Formula mass of CaCO3 = = 100 g/mol
Formula mass of CaO = = 56 g/mol

32. X
CaCO3  CaO + CO2 56
24.8/100 = X/ X = 24.8/100 x 56
X = 13.9 g
% YIELD = 10.6/13.9 X 100 = 76.3%

33. READ UNIT 12.3 IN YOUR TEXT, PP 368 – 375
ON PAGES , WORK:
# 41
#42
#47
#48
EXTRA CREDIT: #60 ON P SHOW ALL STEPS AND BE ABLE TO EXPLAIN.