Presentation is loading. Please wait.

Presentation is loading. Please wait.

Stoichiometry Chapter 12  Turn in:  Forensic Farming Reading  Fill Out Goal Sheet  Our Plan:  Begin Lab  Daily Challenge – Smore’s Stoichiometry.

Similar presentations


Presentation on theme: "Stoichiometry Chapter 12  Turn in:  Forensic Farming Reading  Fill Out Goal Sheet  Our Plan:  Begin Lab  Daily Challenge – Smore’s Stoichiometry."— Presentation transcript:

1

2 Stoichiometry Chapter 12

3  Turn in:  Forensic Farming Reading  Fill Out Goal Sheet  Our Plan:  Begin Lab  Daily Challenge – Smore’s Stoichiometry  Notes – Stoichiometry Conversions  Worksheet #1 & Worksheet #2  Wrap Up - Pyramid  Homework (Write in Planner):  WS#1 & WS#2 Due Monday

4

5  Follow the procedure for Day 1 and begin recording observations  Use the internet to answer the Pre-Lab Questions

6

7 French Toast Recipe:  4 slices of bread  ¼ cup milk  1 egg  1 tsp. vanilla 1.If you were making French Toast for 20 people how many eggs would you need?

8  4 slices of bread  ¼ cup milk  1 egg  1 tsp. vanilla 2.If you were making French toast for 12 people how many teaspoons of vanilla would you use?

9  4 slices of bread  ¼ cup milk  1 egg  1 tsp. vanilla 3.If you have 1 cup of milk, how many slices of French Toast can you make?

10  Stoy-key-ahm-a-tree

11  Comes from the combination of the Greek words stoikheioin, meaning “element”, and metron, meaning “to measure”

12  The study of the quantitative, or measurable, relationships in a chemical reaction.

13  Using stoichiometry, you can determine the quantities of reactants & products in a reaction from the BALANCED equation.  Like the Recipe!

14  Do show the number of moles of each substance involved in the rxn.  Do Not indicate the actual number of grams of the substance.

15 N 2 + 3H 2 --> 2NH 3 Atoms N Atoms H 2 6 On both the reactant and product side!

16 N 2 + 3H 2 --> 2NH 3 Molecule N 2 Molecules H 2 Molecules NH

17 N 2 + 3H 2 --> 2NH 3 Mole N 2 Moles H 2 Moles NH

18 N 2 + 3H 2 --> 2NH 3 Grams N 2 Grams H 2 Grams NH

19  Which were conserved?  Atoms and Mass, but NOT Moles and Molecules!

20  A conversion factor that relates the amounts in moles of any 2 substances involved in a chemical reaction.

21 1 mole N 2 3 moles H 2 2 moles NH 3 1 mole N 2 N 2 + 3H 2 --> 2NH 3

22 Molecules Mass Volume Moles Given Moles Unknown Molecules Mass Volume Molar Ratio Equation x Molar Mass PT Molar Mass PT 22.4 L/mole

23  The conversion of moles of one type of substance to moles of another type of substance.

24 Molar Ratio Equation Moles Given Moles Unknown

25  H 2 + I 2 --> 2HI  If 4 moles of H 2 react, how many moles of HI will form?

26 4 moles H2X 2 moles HI 1 mole H2 =8 moles HI H 2 + I 2 2HI H 2 + I 2 --> 2HI

27  If 0.8 moles of HI form, how many moles of I 2 were used in the rxn?

28 0.8 moles HI X 2 moles HI 1 mole I 2 = 0.4 moles I 2 H 2 + I 2 2HI H 2 + I 2 --> 2HI

29 2H 2 + O 2 --> 2H 2 O 1.If 1.3 moles of H 2 O form, how many moles of O 2 were used in the rxn? (0.65 moles) 2.If 0.21 moles of H 2 react, how many moles of H 2 O will form? (0.21 moles)

30 1.3 moles H2OX 1 moles O2 2 moles H2O = 0.65 moles O2

31 0.21 moles H 2 X 2 moles H 2 O 2 moles H 2 = 0.21 moles H 2 O

32  Given the moles of one substance & asked to determine the mass of another substance.

33 MolesGiven molarratioequation MolesUnknown molarmassPT Mass Unknown Unknown

34 C 6 H 12 O 6 +6O 2 --> 6CO 2 + 6H 2 O WWhat mass of sucrose was used if 3 moles of water formed?

35 3 moles H 2 O X 1 mole C 6 H 12 O 6 6 moles H 2 O = 90g C 6 H 12 O 6 X 180g C 6 H 12 O 6 1 mole C 6 H 12 O 6 C 6 H 12 O 6 +6O 2 6CO 2 + 6H 2 O C 6 H 12 O 6 +6O 2 --> 6CO 2 + 6H 2 O

36  What mass of carbon dioxide was used to form 5.0 moles of water?

37 5.0 moles H 2 O X 6 moles CO 2 6 moles H 2 O = 220 g CO 2 X 44g CO 2 1 mole CO 2 C 6 H 12 O 6 +6O 2 6CO 2 + 6H 2 O C 6 H 12 O 6 +6O 2 --> 6CO 2 + 6H 2 O

38 N 2 O 5 + H 2 O --> 2HNO 3 1.What mass of water was used to form 2.5 moles of HNO 3 ? (23 g H 2 O) 2.What mass of HNO 3 will form from moles of N 2 O 5 ? (1512 g HNO 3 )

39 2.5 moles HNO 3 X 2 moles HNO 3 1 mole H 2 O = 23 g. H 2 O X 18g. H 2 O 1 mole H 2 O

40 12.00 moles N2O5 X 2 moles HNO3 1 mole N2O5 =1512 g. HNO3 X 63g. HNO3 1 mole HNO3

41  Given mass of one substance and asked to find moles of another substance.

42 MolesGiven molarratioequation MolesUnknown molarmassPT MassGiven

43 2NO + 3H 2 --> 2NH 3 + O 2  How many moles of NO were used to form 824g. NH 3 ?

44 48.5 moles NO X 2 moles NO 2 moles NH 3 = 824g. NH 3 X 17g. NH 3 1 mole NH 3 2NO + 3H 2 2NH 3 + O 2 2NO + 3H 2 --> 2NH 3 + O 2

45 29.3 moles H 2 X 3 moles H 2 1 mole O 2 = 312g. O 2 X 32g. O 2 1 mole O 2 2NO + 3H 2 2NH 3 + O 2 2NO + 3H 2 --> 2NH 3 + O 2

46 Pencil Cougars Chemistry Purple Safety ShowerMole $100,000 Pyramid

47 Stoichiometry CoefficientMolar Mass Periodic Table BalanceRatio $100,000 Pyramid

48  Turn in:  Worksheet #1 & Worksheet #2  Our Plan:  Conversions Table Review  Notes – Mass to Mass Conversions  Worksheet #3  Homework (Write in Planner):  Worksheet #3 due next class

49 Molecules Mass Volume Moles Given Moles Unknown Molecules Mass Volume Molar Ratio Equation x Molar Mass PT Molar Mass PT 22.4 L/mole

50  As a group of 3 or 4, complete the problems on the team conversion review. Each student should have a different colored writing utensil and you should take turns doing the work.

51 Molecules Mass Volume Moles Given Moles Unknown Molecules Mass Volume Molar Ratio Equation x Molar Mass PT Molar Mass PT 22.4 L/mole

52  Given the mass of one substance and asked to determine the mass of another substance.

53 MolesGiven molarratioequation MolesUnknown molarmassPT Mass Given Given molarmassPT Mass Unkn Unkn

54  2Na + Cl 2 -->2NaCl  Find the mass of table salt produced from 18.6g Na.

55 47.3g NaCl X 2 moles NaCl 2 mole Na = 18.6g. Na X 23g. Na 1 mole Na 58.5g NaCl 1 mole NaCl X 2Na + Cl 2 --> 2NaCl

56  2Al + Fe 2 O 3 --> Al 2 O 3 + 2Fe  Find the mass of aluminum oxide produced from 2.3g of aluminum.

57 4.3 g Al 2 O 3 X 1 mole Al 2 O 3 2 mole Al = 2.3g. Al X 27g. Al 1 mole Al 102g Al 2 O 3 1 mole Al 2 O 3 X 2Al + Fe 2 O 3 --> Al 2 O 3 + 2Fe

58 CH 4 + 2O 2 --> CO 2 + 2H 2 O  Methane burns in air by the following reaction: CH 4 + 2O 2 --> CO 2 + 2H 2 O  What mass of water is produced by burning 500. g of methane?

59 1,130 g H 2 O X 2 mole H 2 O 1 mole CH 4 = 500.g CH 4 X 16g CH 4 1 mole CH 4 18g H 2 O 1 mole H 2 O X

60  Complete Worksheet #3 by next class!

61  Turn in:  Worksheet #3  Our Plan:  Questions on WS 1-3  Stoichiometry Team Review  Quiz WS 1 – 3  Lab Day 2  Homework (Write in Planner):  Finish missing worksheets

62 Molecules Mass Volume Moles Given Moles Unknown Molecules Mass Volume Molar Ratio Equation x Molar Mass PT Molar Mass PT 22.4 L/mole

63 ,

64

65  Turn in:  Any Missing Worksheets  Our Plan:  Lab Day 3  Turn in Lab Today  Homework (Write in Planner):  Lab, if not done

66 #1 – 1000 mg = 1 g, 1000 mL = 1 L #3 – Convert both the 1g AgNO 3 and your g of Cu to g of Ag. The limiting reactant is the reactant that produces the least product. You will do 2 three step conversions to find the answer. # 5 & #6 – Percent yield = (actual/theoretical) x 100 WASH ALL SUPPLIES, RETURN TO SIDE TABLE, WIPE DOWN COUNTER, WASH HANDS!

67  Clicker Review Clicker Review

68  Turn in:  Metal in Water Lab  Our Plan:  Relay Race Review  Daily Challenge  Notes – Stoichiometry of Gases and Molecules  Worksheet #4  Homework (Write in Planner):  WS#4 due next class

69 1. How many liters are in 1 mole of a gas? 2. How many molecules are in 1 mole of any compound? 3. For the reaction below, what mass of water can be produced from 1.5 moles of hydrogen? (27g) 2H 2 + O 2 --> 2H 2 O

70 Molecules Mass Volume Moles Given Moles Unknown Molecules Mass Volume Molar Ratio Equation x Molar Mass PT Molar Mass PT 22.4 L/mole

71  Given the volume of one substance and asked to determine the volume of another substance.

72 MolesGiven molarratioequation MolesUnknown 22.4 L Vol Given Given 22.4 L Vol Unkn Unkn

73  Conversion factor: 22.4 L/mole

74  You can only use the conversion factor 22.4 under very specific conditions.  It must be a gas and it must be at STP (Standard Temperature and Pressure)  STP is 0 degrees Celsius and 1 atm

75  How many moles of Helium are in a 3.7 L balloon at STP?

76 3.7 L x 1 mole = 0.17 mole 22.4 L

77 2 CO + O 2 --> 2 CO 2  What volume of carbon dioxide will be produced from 2.6 L of oxygen at STP?

78 Molecules Mass Volume Moles Given Moles Unknown Molecules Mass Volume Molar Ratio Equation x Molar Mass PT Molar Mass PT 22.4 L/mole

79 2.6 L O 2 X1 mole O 2 X 22.4 L O 2 2 mole CO 2 X 1 mole O L CO 2 = 1 mole CO L CO 2 2 CO + O 2 --> 2 CO 2

80 UUsing the same equation, what volume of carbon dioxide will be produced at STP from 45.3 grams of carbon monoxide?

81 Molecules Mass Volume Moles Given Moles Unknown Molecules Mass Volume Molar Ratio Equation x Molar Mass PT Molar Mass PT 22.4 L/mole

82 45.3 g CO X1 mole CO X 28 g CO 2 mole CO 2 X 2 mole CO 22.4 L CO 2 = 1 mole CO L CO 2 2 CO + O 2 --> 2 CO 2

83 (4.91 L)  Using the same equation, what volume of oxygen is used to produce 19.3 grams of carbon dioxide at STP? (4.91 L)

84 19.3 g CO 2 X1 mole CO 2 X 44 g CO 2 1 mole O 2 X 2 mole CO L O 2 = 1 mole O L O 2 2 CO + O 2 --> 2 CO 2

85  You can do the same type of problems using molecules. Just use Avogadro’s Number!

86  In the reaction: 2H 2 + O 2 --> 2H 2 O if you have 0.50 grams of hydrogen, how many molecules of water will form?

87 Molecules Mass Volume Moles Given Moles Unknown Molecules Mass Volume Molar Ratio Equation x Molar Mass PT Molar Mass PT 22.4 L/mole

88 1.5 x molecules H 2 O X 2 mole H 2 O 2 mole H 2 = 0.50 g H 2 X 2 g H 2 1 mole H x molecules 1 mole H 2 O X 2H 2 + O 2 --> 2H 2 O

89 CCooking PPlanning outcomes of reactions RRace cars/ mechanics

90 TTreating an upset stomach AAirbags AArtists

91 LLaunching space shuttle BBaking bread EEngineering

92  Complete Worksheet #4 by next class period!

93  Turn in:  Get Worksheet #4 Out to Check  Our Plan:  Balloon Races  Limiting Reactant Notes  Finish Balloon Races Handout  Worksheet #5  Chemistry Cartoon  Homework (Write in Planner):  Worksheet 5 & Cartoon Due Next Class

94 Molecules Mass Volume Moles Given Moles Unknown Molecules Mass Volume Molar Ratio Equation x Molar Mass PT Molar Mass PT 22.4 L/mole Pre Lab Review – The balanced equation for the reaction of baking soda and vinegar is: HC 2 H 3 O 2 + NaHCO 3 --> NaC 2 H 3 O 2 + CO 2 + H 2 O. How many liters of carbon dioxide can form from 3.00 g of baking soda (NaHCO 3 )? Show your work at the top of your lab.

95

96  What did you predict would happen?  What did happen? Why?

97 Smore’s Recipe 2 Graham Crackers 3 Pieces of Chocolate 1 Marshmallow

98 1. If you have 4 graham crackers, 9 pieces of chocolate, and 7 marshmallows, how many smores can you make?

99 2. Which ingredient determined this?

100 3. If you have 22 graham crackers, 27 pieces of chocolate, & 13 marshmallows, how many smores can you make?

101 4. Which ingredient determined this?

102  Even in abundance, the reactants must still combine in proportion  They follow stoichiometry.

103  Is the reactant that limits the amount of product formed in a chemical equation.

104  Is the substance that is not used up completely in a reaction.

105  The quantities of products formed in a reaction are always determined by the quantity of the limiting reactant.

106 1.Write a balanced equation 2.Convert the mass of reactants to the mass of the product.

107 3.Whichever produces less product is the limiting reactant.

108  Identify the L.R when 10.0 g of water react with 4.50 g of sodium to produce sodium hydroxide & hydrogen gas.

109 2H 2 O + 2Na --> 2NaOH + H g NaOH X 2 mol NaOH 2 mol H 2 O = 10.0 g H 2 O X 18g H 2 O 1 mol H 2 O 40g NaOH 1 mol NaOH X 7.83 g NaOH X 2 mol NaOH 2 mol Na = 4.50 g Na X 23g Na 1 mol Na 40g NaOH 1 mol NaOH X Na= L.R.

110  Identify the L.R when 8.90 g HF are combined with 14.56g SiO 2 in the following reaction: SiO 2 + 4HF --> SiF 4 + 2H 2 O

111 4.0 g H 2 O X 4 mol HF 2 mol H 2 O = 8.9 g HF X 20g HF 1 mol HF 18g H 2 O 1 mol H 2 O X g H 2 O X 2 mol H 2 O 1 mol SiO 2 = 14.50g SiO 2 X 60g SiO 2 1 mol SiO 2 18g H 2 O 1 mol H 2 O X HF= L.R. SiO 2 + 4HF -->SiF 4 + 2H 2 O

112  Finish Balloon Races Questions  Worksheet #5  Chemistry Cartoon

113

114

115

116

117  Turn in:  Get Worksheet #5 Out to Check  Turn in Chemistry Cartoon - Basket  Our Plan:  L.R. Review Problem  Percent Yield Notes  Worksheet #6  Homework (Write in Planner):  Worksheet #6 Due Tuesday/Wednesday

118 Acrylonitrile, C 3 H 3 N, is an important ingredient in the production of various fibers and plastics. Acrylonitrile is produced from the following reaction: C 3 H 6 + NH 3 + O 2 --> C 3 H 3 N + H 2 O If 850 g of C 3 H 6 is mixed with 300. g of NH 3 and unlimited O 2, to produce 850. g of acrylonitrile, what is the percent yield? You must first balance the equation. 91 %

119 1. Babe Ruth played in 2503 baseball games in his career. He had 2873 hits in 8399 at bats what was his batting average? 

120 2. Hank Aaron played in 3298 baseball games in his career. He had 3771 hits in 12,364 at bats. What was his batting average? 

121  Theoretical Yield is based on calculations

122  Actual Yield is based on the actual chemical reaction (the lab)

123  % yield = Actual yield X 100 Theoretical yield

124 1.Write equation 2.Calculate the mass of the product that should have been formed. (theoretical yield)

125 3.Plug numbers into the equation

126  Determine the % yield for the reaction between 2.80 g Al(NO 3 ) 3 & excess NaOH if 0.966g Al(OH) 3 is recovered.

127 WWrite the equation: Al(NO3)3 + 3NaOH --> Al(OH)3 + 3NaNO3

128  Calculate theoretical yield 1.03 g Al(OH) 3 X 1 mol Al(OH) 3 = 2.80 g Al(NO 3 ) 3 X 213g Al(NO 3 ) 3 1 mol Al(NO 3 ) 3 78g Al(OH) 3 X 1 mol Al(NO 3 ) 3 1 mol Al(OH) 3 Al(NO 3 ) 3 + 3NaOH --> Al(OH) 3 + 3NaNO 3

129 PPlug numbers into equation % yield= 0.966g 1.03g X 100 =93.8%

130  Determine the percent yield for the reaction between 15.0 g N 2 & excess H 2 if 10.5g NH 3 is produced.

131 18.2 g NH 3 X 1 mol N 2 2 mol NH 3 = 15.0g N 2 X 28g N 2 1 mol N 2 17g NH 3 1 mol NH 3 X %yield = 10.5g NH g NH 3 X 100=57.7%

132  You have to find the limiting reactant first.  Calculate amount of product for both reactants & whichever is less, use as theoretical yield.

133  Complete Worksheet #6 by next class!

134  Turn in:  Worksheet #6  Our Plan:  Stoichiometry Lab Test  Test Review  Homework (Write in Planner):  Test Review due next class  UNIT 8 TEST NEXT CLASS!

135  Turn in:  Get out Test Review  Our Plan:  Questions on Test Review?  Test  Larry the Lawnchair Guy Reading  Homework (Write in Planner):  Larry Reading due next class

136  Given the following reaction: C 3 H 8 + 5O > 3CO 2 + 4H 2 O If you start with 14.8 g of C 3 H 8 and 3.44 g of O 2, determine the limiting reagent  Given the following equation: H 3 PO KOH > K 3 PO H 2 O If 49.0 g of H 3 PO 4 is reacted with excess KOH, determine the percent yield of K 3 PO 4 if you isolate 49.0 g of K 3 PO 4.

137  Given the following equation: Al(OH) 3 +3 HCl --> AlCl H 2 O If you start with 50.3 g of Al(OH) 3 and you isolate 39.5 g of AlCl 3, what is the percent yield?  Given the following equation: H 2 SO 4 + Ba(OH) > BaSO 4 + H 2 O If 98.0 g of H 2 SO 4 is reacted with excess Ba(OH) 2, determine the percent yield of if g of BaSO 4 is formed.

138  Km7aUjgc Km7aUjgc


Download ppt "Stoichiometry Chapter 12  Turn in:  Forensic Farming Reading  Fill Out Goal Sheet  Our Plan:  Begin Lab  Daily Challenge – Smore’s Stoichiometry."

Similar presentations


Ads by Google