CHEMICAL REACTIONS

Page 12

44. BALANCE THESE EQUATIONS AND CLASSIFY AS BEING A SYNTHESIS, DECOMPOSITION, SINGLE OR DOUBLE REPLACEMENT, OR COMBUSTION REACTION:

2K + Cl2 2KCl Type: Synthesis

2AlBr3 + 3 Na2(CO3) Al2(CO3)3 +6 NaBr Type: Double replacement

2C2H6 + 7O2 4 CO2 +6 H2O Type:Combustion

3Cu2(CrO4) + 2 Fe Fe2(CrO4)3 + 6Cu

Type: Single replacement

Mg(CO3) MgO + CO2 Type: decomposition

3H2 + Fe2S3 3 H2S + 2 Fe Type: Single replacement

45. IDENTIFY THE SOLUBILITY OF EACH COMPOUND IN THE DOUBLE REPLACEMENT REACTION ABOVE

2AlBr3 (aq)+ 3 Na2(CO3) (aq) Al2(CO3)3 (s) +6 NaBr(aq)

Precipitate!(The insoluble product)

46. WHICH SINGLE REPLACEMENT REACTION ABOVE WILL NOT OCCUR? 3H2 + Fe2S3 3 H2S + 2 Fe will not occur.

Iron is more reactive than hydrogen. (Refer to the activity series on the back of your periodic table). The more reactive element will be in the compound.

47A. WRITE THE SKELETON EQUATION THEN BALANCE Copper(II) chloride reacts with iron to

produce iron(III) chloride and copper metal.

Skeleton: CuCl2 + Fe FeCl3 +Cu

Balanced: 3CuCl2 + 2Fe 2FeCl3 +3Cu

Type of reaction : Single replacement

47B. WRITE THE SKELETON EQUATION THEN BALANCE Hydrogen gas and bromine liquid react

to yield hydrogen bromide.

Skeleton: H2 + Br2 HBr

Balanced: H2 + Br2 2HBr

Type of reaction : Synthesis

47C. WRITE THE SKELETON EQUATION THEN BALANCE Carbon tetrahydride reacts with oxygen

to produce carbon dioxide and water vapor.

Skeleton: CH4 + O2 CO2+ H2O

Balanced: CH4 + 2O2 CO2 + 2H2O

Type of reaction : Combustion

STOICHIOMETRY Pages 13 and 14

48. HOW MANY MOLES OF CALCIUM CHLORIDE CAN BE THEORETICALLY YIELDED FROM 1.53 MOLES OF HYDROGEN CHLORIDE?

___HCl + ___Ca(OH)2 ___CaCl2 + ___H2O

Balance: 2HCl + 1Ca(OH) 2 1CaCl2 +2H2O

Use the mole ratio from the balanced equation to convert from moles of HCl to moles of CaCl2 :

2HCl + 1Ca(OH) 2 1CaCl2 +2H2O

1.53 mol HCl 1 mol CaCl2 = .765 mol CaCl2 2 mol HCl

49. HOW MANY GRAMS OF IRON(III) BROMIDE WILL BE THEORETICALLY YIELDED FROM THE REACTION OF 2.05 MOLES OF IRON?

___HBr + ___Fe ___FeBr3 + ___H2

Balance: 6HBr + 2Fe 2FeBr3 + 3H2 Use the mole ratio from the balanced equation

to convert from moles of Fe to moles of FeBr3: Use molar mass to convert between moles &

grams. 6HBr + 2Fe 2FeBr3 + 3H2

2.05 mol Fe 2 mole FeBr3 295.557g = 606g FeBr3

2 mole Fe 1 mole FeBr3 Molar mass of FeBr3

50. IN A REACTION OF SODIUM WITH CHLORINE, 2.30 GRAMS OF CHLORINE ARE CONSUMED. HOW MANY GRAMS OF PRODUCT CAN BE THEORETICALLY YIELDED?

___Na + ___Cl2 ___NaCl

Balance: 2Na + 1Cl2 2NaCl

Use the mole ratio from the balanced equation to convert from moles of Na to moles of Cl2:

Use molar mass to convert between moles & grams.

2Na + 1Cl2 2NaCl

2.3g Na 1 mol Na 2 mol NaCl 58.44g NaCl = 5.8g NaCl

22.990g Na 2 mol Na 1 mol NaClMolar mass of NaCl

Molar mass of Na

51. NITROGEN GAS CAN BE PREPARED BY PASSING GASEOUS AMMONIA (NH3) OVER SOLID COPPER(II) OXIDE AT HIGHTEMPERATURES. THE OTHER PRODUCTS OF THE REACTION ARE SOLID COPPER AND WATER VAPOR.

A. If a sample containing 18.1 grams of NH3 reacted with 90.4 grams of copper(II) oxide, which is the limiting reactant?

2NH3 +3CuO 1N2 + 3Cu + 2H2O

18.1g NH3 1 mol NH3 1 mol N2 28.014 g N2 = 14.9 g N2

17.031g NH3 2 mol NH3 1 mol N2

90.4g CuO 1 mol CuO 1 mol N2 28.014 g N2 = 10.6 g N2

79.545g CuO 3 mol CuO 1 mol N2

A. CuO is the limiting reactant.

B. How many grams of N2 will theoretically be formed? 10.6g N2

52. A STUDENT ADDS 200.0G OF C7H6O3 TO AN EXCESS OF C4H6O3, THIS PRODUCES C2H4O2 AND C2H4O2. CALCULATETHE PERCENT YIELD IF 231 G OF ASPIRIN (C9H8O4) IS PRODUCED.

1C7H6O3 + 1C4H6O3 1C9H8O4 + 1C2H4O2

200.0g C7H6O3 1 mol C7H6O3 1 mol C9H8O4 180.069g C9H8O4 =

138.052g C7H6O3 1 mol C7H6O3 1 mol C9H8O4

Answer = 260.9g theoretical

Actual x 100 = % yield

Theoretical

231 x100 = 88.5%

260.9

53. BENZENE (C6H6) REACTS WITH BROMINE TO FORM BROMOBENZENE (C6H5BR ) IN THE REACTION BELOW. A. WHAT IS THE THEORETICAL YIELD OF BROMOBENZENE IN THIS REACTION WHEN 30.0 GRAMS OF BENZENE REACTS WITH 65.0 GRAMS OF BROMINE?

1C6H6 + 1Br2 1C6H5Br + 1HBr

30.0g C6H6 1 mol C6H6 1 molC6H5Br 156.95g C6H5Br = 60.32gC6H5Br

78.054g C6H6 1 mol C6H6 1 mol C6H5Br

65.0g Br2 1 mol Br2 1 mol C6H5Br 156.95g C6H5Br = 63.8 C6H5Br

159.808g Br2 1 mol Br2 1 mol C6H5Br

C6H6 was the limiting reactant, so only 60.32 g of C6H5Br would be theoretically produced.

53.B IF THE ACTUAL YIELD OF BROMOBENZENE WAS 42.3 GRAMS, WHAT WAS THE PERCENT YIELD?C6H6 was the limiting reactant, so only 60.32 g of

C6H5Br would be theoretically produced.

Actual x 100 = % yield

Theoretical

42.3 x100 = 65.82%

60.32

STATES OF MATTER Pages 15,16,17, 18

54. CLASSIFICATION OF MATTER

Matter

Pure Substances Mixtures

Elements CompoundsHomogeneous

MixturesHeterogeneous

Mixtures

55. DEFINE:

a. matter – anything that has mass or takes up space

b. physical property – property of matter that can be observed or measured without changing the substance

Examples: color, density, mass, volume

55. DEFINE: c. extensive physical property –

depends on amount of substance present

Examples: mass, volume, length

d. intensive physical property – does not depends on the amount of a substance

Examples: density, color, melting point, boiling point

55. DEFINEe. Chemical property- used to describe

abilityExamples: reactivity, flammability,

separating mixtures

f. Physical change- alters the appearance but does not change the composition of the substance.

Examples: phase change, separating mixtures, dissolving, evaporating

56. WHICH PHYSICAL SEAPARATION TECHNIQUE? a. Peas and carrots- chromatography to

separate by color. b. charcoal powder and iron powder-

chromatography to separate by color. c. salt water- salt dissolves in water so

you would use crystallization d. pigments in green food coloring- if

the pigments are dissolved use crystallization, if not dissolved use filtration

e. rubbing alcohol and water- distillation to separate by boiling points

57 AND 58 57. Define chemical change – change

occurs when one or more substances undergoes a chemical reaction to form a new substance

Examples: cooking, combustion, oxidation, fizzes

58. List AND EXPLAIN the four indicators of a chemical change.

a.Color change c. gas evolution

b. formation of a precipitate d. odor

59. INDICATE ENDOTHERMIC OR EXOTHERMIC a. A student pours hydrochloric acid into

a test tube containing a white, crystalline powder. The mixture begins to bubble and the test tube begins to feel cold. ENDOTHERMIC

b. A student pours HCl at 25.2 C into a test tube containing a small metal strip. The mixture begins to fizz and the temperature of the mixture rises to 38.6 C.EXOTHERMIC

60. STATES OF MATTERSolids Liquids Gases

Compressibility

Incompressible

Incompressible

Compressible

Structure Tightly packed

Loosely packed particles

No attraction between particle

Motion Particles vibrate

Ability to flow Move freely ability to flow

Shape Fixed shape Takes shape of container

Spread out in container

Volume Fixed shape Fixed volume Depends on size of container

61.LIST THE SIX PHASE CHANGES, DEFINE THEM, AND GIVE AT LEAST ONE EXAMPLE OF EACH:

A. melting- solid to liquidB. Freezing- liquid to solid C. Vaporization- liquid to gasD. condensation- gas to liquidE. Sublimation- solid to gas F. deposition- gas to solid

62.

Solid

Liquid

Gas

Triple point- point where all 3 phases coexist

critical point- anything above this point will be a gas

63. DEFINE a. Pure substance – uniform

unchanging composition, ex: elements and compounds

b. Element- single type of atom ex: gold (Au), Hydrogen (H)

c. Compound- more than one element combined ex: NaCl

63. DEFINE d. Mixture- combination of two or

more pure substances ex: salt water

e. Homogeneous mixture- constant composition throughout and are always in one phase

f. Solution- homogeneous mixture

63. DEFINE g. Heterogeneous mixture- mixtures do

not blend together smoothly and the individual substances remain distinct ex: colloid, suspension

h. Colloid- one substance is suspended evenly throughout another substance ex: milk, fog, jello

i. Suspension-large substance particles are suspended in another substance.

Ex: muddy water, paint

REFER TO GRAPH FOR QUESTIONS 64- 69

64-66 USE SOLUBILITY GRAPH64. 85 grams

65. 130 grams

66. 50 oC

SOLUTIONSPages 19,20

REFER TO GRAPH FOR QUESTIONS 64- 69

67. unsaturated solution

68. 30g will dissolve, 20 grams will remain at the bottom of the beaker

69. 25 more grams will dissolve

70. MOLARITY FORMULA

Molarity = moles of soluteLiters of solution

Molarity units = M (concentration)You may have to covert mL to LGiven mL 1 L

1000 mLYou may have to covert grams to

moles in order to solve for MolarityGiven grams 1 mole

molar mass

71. WHAT FORMULA IS USED FOR DILUTING A CONCENTRATED SOLUTION TO A GIVEN MOLARITY?

M1V1= M2V2

M= Molarity

V= Volume

72. DETERMINE THE MOLARITY OF 500 ML OF A SOLUTION CONTAINING 7.20 G OF SODIUM ACETATE Molarity = moles of solute

Liters of solution Convert Grams moles

7.20g 1 mole = .087 moles

83 g Convert mL L

500mL 1L = .500L

1000 mL Plug into equation:

Molarity = .087 = .18M

.500

73. HOW MANY MOLES OF CALCIUM CHLORIDE ARE IN 1.35 L OF A 2.5 M SOLUTION OF CALCIUM CHLORIDE?

Molarity = moles of solute Liters of solution

Plug into equation:

2.5 = X = 3.375 mol

1.35

74. HOW MANY MILLILITERS OF A 5.0 M H2SO4 STOCK SOLUTION WOULD YOU NEED TO PREPARE 100.0 ML OF 0.25 MH2SO4??

Formula: M1V1=M2V2 M = Molarity, V = volume Solving for V1

(5.0M)(V1)= (.25M)(100mL)

(5.0M)(V1)= 25

(V1)= 5mL

Multiply .25 x 100

Divide both sides by 5.0 to get (V1) by itself

75. IF YOU DILUTE 20.0 ML OF A 3.5 M SOLUTION TO MAKE 100.0 ML OF SOLUTION, WHAT IS THE MOLARITY OF THEDILUTE SOLUTION?

Formula: M1V1=M2V2 M = Molarity, V = volume Solving for M2

(3.5M)(20mL)= (M2)(100mL)

70 = (M2)(100mL)

.7M= (M2)

Multiply 3.5 x 20

Divide both sides by 100 to get (M2) by itself