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

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

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

CALCULATE THE MOLAR MASS OF ZnSO4.

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

HOW MANY MOLES OF K2SO4 ARE IN 20 GRAMS?

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 = 0.115 mole

HOW MANY MOLECULES OF C2H5OH ARE IN 2.5 MOLES?

HOW MANY MOLECULES OF C2H5OH ARE IN 2.5 MOLES? 1 mole = 6.023 x 1023 molecules # molecules = 2.5 moles x 6.023 x 1023 molecules/mole = 1.51 x 1024 molecules

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

HOW MANY GRAMS OF PbI2 ARE IN 8.4 X 1022 MOLECULES? # moles = 8.4 x 1022 / 6.023 x 1023 = 1.39 x 10-1 = 0.139 moles Molar mass of PbI2 = 207 + 2(127) = 461 g/mole # grams = 461 g/mole x 0.139 moles = 64 g

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

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 C4H10 + 13 O2 → 8 CO2 + 10 H2O 2 Fe + 3 Cl2  2 FeCl3 Cu + 2 AgNO3  Cu(NO3)2 + 2 Ag 3 Pb(NO3)2 + 2 AlCl3  2 Al(NO3)3 + 3 PbCl2 2 KClO3  2 KCl + 3 O2

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?

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

WHAT IS THE LAW OF CONSERVATION OF MASS?

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

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

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.

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

How many grams of ammonia (NH3) would be required to make 1 kilogram of ammonium sulfate? X 1000 g 2 NH3 + H2SO4  (NH4)2SO4 34 128 X / 34 = 1000 / 128 X = (1000 x 34) / 128 = 265.6 g Note: molar mass of ammonium sulfate = 128 g/mole molar mass of ammonia = 17 g/mole

How many moles of iron, Fe, would be required to make 40 grams of hydrogen, H2? 2 Fe + 6 HCl  2 FeCl3 + 3 H2

How many moles of iron, Fe, would be required to make 40 grams of hydrogen, H2? 2 Fe + 6 HCl  2 FeCl3 + 3 H2 Molar mass of H2 = 2 g/mole # moles H2 = 40 g/2 g/mole = 20 moles X 20 2 3 X/2 = 20/3 so X = 20/3 x 2 = 13.3 moles

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?

EMPIRICAL FORMULA: CuCl2 Molar mass = 63.5 + 2(35.5) = 134.5 g/mole % Cu = (63.5 / 134.5) x 100 = 47.2 % % Cl = (71 / 134.5) x 100 = 52.8 %

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.

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 = 0.779 moles 1.26 X 2 Cu + S  Cu2S 1.26/2 = X/1 x = 0.63 2 1 So, it would take 0.63 moles of S to react with 1.26 moles Cu. You have 0.779 moles S, so Cu is the limiting reagent and S is in excess.

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.

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

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 = 40 + 12 + 48 = 100 g/mol Formula mass of CaO = 40 + 16 = 56 g/mol

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

READ UNIT 12.3 IN YOUR TEXT, PP 368 – 375 ON PAGES 379-380, WORK: # 41 #42 #47 #48 EXTRA CREDIT: #60 ON P 381. SHOW ALL STEPS AND BE ABLE TO EXPLAIN.