Presentation is loading. Please wait.

Presentation is loading. Please wait.

Stoichiometry for Strugglers Stoichiometry for Strugglers Website: allenisd.org/Page/49042 by Daniel Haradem & Kathy Vondracek of Allen ISD for CAST2016.

Published byEaster Bond Modified over 8 years ago

Similar presentations

Presentation on theme: "Stoichiometry for Strugglers Stoichiometry for Strugglers Website: allenisd.org/Page/49042 by Daniel Haradem & Kathy Vondracek of Allen ISD for CAST2016."— Presentation transcript:

1 Stoichiometry for Strugglers Stoichiometry for Strugglers Website: allenisd.org/Page/49042 by Daniel Haradem & Kathy Vondracek of Allen ISD for CAST2016 in San Antonio, TX Workshop 1041

2 Tricks to Trade -- The scribble (It goes underneath the # that you are starting with) -- Green and Pink Highlighters (Go and Stop) -- The Tape (Conversation factors based off 1 base unit) -- Swish/Swish song (used for cancelling units) -- Showing your work(use parenthesis and multiply signs) -- Calculator (alternating multiply and divide)

3 Pressure Conversion with Pressure Tape Pressure Conversion with Pressure Tape 1 atm = 760 mm Hg = 760 torr =101.33 KPa = 14.7 psi Your car tires require 35.0 psi to run optimally. How many atmosphere is this? 2.38 = x atm psi atm 14.7 1 35.0 psi

4 The Metric Tape. -- Based of 1 base unit. -- You can create you type based off your needs 0.001 Kg = 1 g = 10 dg = 100 cg = 1,000 mg -- 4 types of metric tape (mass, volume, distance and generic) 1 x 10 -3 Km = 1 m = 1 x 10 2 cm =1 x 10 3 mm = 1 x 10 6 µm = 1 x 10 9 nm 1 x 10 -3 KL = 1 L = 1 x 10 2 cL =1 x 10 3 mL = 1 x 10 6 µL = 1 x 10 9 nL 1 x 10 -3 Kg = 1 g = 1 x 10 2 cg =1 x 10 3 mg = 1 x 10 6 µg = 1 x 10 9 ng 1 x 10 -3 K_ = 1 _ = 1 x 10 2 c_ =1 x 10 3 m_ = 1 x 10 6 µ_ = 1 x 10 9 n_

5 Metric Conversion with Metric Tape 1 x 10 -3 Km = 1 m = 1 x 10 2 cm =1 x 10 3 mm = 1 x 10 6 µm = 1 x 10 9 nm A typical atom is approximately 0.100 nanometers across. What is this value in centimeters? 0.100 nm 1.00 x 10 -8 = x cm m nm cm m 1 x 10 9 1 1 x 10 2 1 x

6 Metric Conversion (Transitive Law) 1 x 10 -3 Km = 1 m = 1 x 10 2 cm =1 x 10 3 mm = 1 x 10 6 µm = 1 x 10 9 nm A typical atom is approximately 0.100 nanometers across. What is this value in centimeters? 0.100 nm 1.00 x 10 -8 = x cm nm cm 1 x 10 9 1 x 10 2

7 Metric Conversion (Proportions) 1 x 10 -3 Km = 1 m = 1 x 10 2 cm =1 x 10 3 mm = 1 x 10 6 µm = 1 x 10 9 nm A typical atom is approximately 0.100 nanometers across. What is this value in centimeters? Cross-multiply and divide Either Commit to Dimensional Analysis or Proportions….. (NOT BOTH) 0.100 nm ??? = cm 1 x 10 9 nm 1 x 10 2 cm Ans: 1.00 x 10 -8 cm

8 6.02 x 10 23 Compound Stoichiometry with Mole Tape 1 mol Liters of gas at STP MM (g) From PT = Atoms Molecules Formula units 22.4 = = Milk of magnesia, Mg(OH) 2 is used to neutralize stomach acid. How many formula units are in 26.89 g of magnesium hydroxide Mg(OH) 2 ? 26.89 g Mg(OH) 2 X mol X 2.775 x 10 23 form.u Mg(OH) 2 g 58.33 MM of Mg(OH) 2 Mg = 1 x 24.31 = 24.31 O = 2 x 16.00 = 32.00 H = 2 x 1.01 = + 2.02 58.33 g/mol Note: This is same method as the pressure and metric problems just different units and numbers = 1 1 6.02 x 10 23 MM form u. mol

9 Chemical Reactions (MOLE RATIO) 4 NH 3 + 5 O 2  6 H 2 O + 4 NO How many moles of NO are produced in the reaction if 15 mol of H 2 O are also produced? 15 mol H 2 O mol NO = mol H 2 O mol NO 10. X The WHAM / WHAM noise is when I placing in the MOLE RATIO. It reminds the students that need to look at the REACTION to get the numbers. 6 4

10 Plan Your Stoichiometry -- Always figure out where you are starting and where you are going. Checklist -- Go to moles (TAPE) -- Use mole ratio (REACTION) -- Go away from moles (TAPE) -- Steps may be deleted based on your start and finish Mole Map --Vertical Arrows are (TAPE) steps Horizontal Arrows are mole ratios (REACTION) steps --Arrows may be deleted based on your start and finish

11 6.02 x 10 23 Reaction Stoichiometry with CHECKLIST 1 mol Liters of gas at STP MM (g) From PT = Atoms Molecules Formula units 22.4 = = 12 g O 2 mol O 2 X = 9.0 g O 2 MM How many grams of NO is produced if 12 g of O 2 is combined with excess ammonia? 4 NH 3 + 5 O 2  6 H 2 O + 4 NO X 32.00 g NO mol NO -- Go to moles (Tape) -- Mole ratio with heat (Reaction) -- Go away from moles (Tape) 12 g O 2 ???g NO X 1 5 1 MM 30.01 4 mol NO g NO

12 6.02 x 10 23 Reaction Stoichiometry with MOLE MAP 1 mol Liters of gas at STP MM (g) From PT = Atoms Molecules Formula units 22.4 = = 12 g O 2 X mol O 2 X = 9.0 g O 2 How many grams of NO is produced if 12 g of O 2 is combined with excess ammonia? 4 NH 3 + 5 O 2  6 H 2 O + 4 NO g NO X 32.00 g NO 12 g O 2 mol O 2 ??? g NO mol NO USE MM of NO USE MOLE RATIO mol NO USE MM of O 2 MM 1 30.01 4 1 MM 5 mol O 2 mol NO

13 Reaction Stoichiometry (CHECKLIST) How many moles of sodium sulfate could potentially be produced byreacting excess sulfuric acid with 120. grams of sodium hydroxide in aneutralization reaction? How many moles of sodium sulfate could potentially be produced by reacting excess sulfuric acid with 120. grams of sodium hydroxide in a neutralization reaction? 2 NaOH + H 2 SO 4  2 H 2 O(s) + Na 2 SO 4 120. g NaOH x x g NaOH mol NaOH mol Na 2 SO 4 = 1 mol MM (g) From PT = 6.02 x 10 23 Atoms Molecules Formula units = = Liters of gas at STP 22.4 -- Go to moles (Tape) -- Mole ratio (Reaction) -- Go away from moles (Tape) 40.00 1.50 120. g NaOH ??mol Na 2 SO 4 mol Na 2 SO 4 MM mole NaOH 1 1 2

14 Reaction Stoichiometry (MOLE MAP) How many moles of sodium sulfate could potentially be produced byreacting excess sulfuric acid with 120. grams of sodium hydroxide in aneutralization reaction? How many moles of sodium sulfate could potentially be produced by reacting excess sulfuric acid with 120. grams of sodium hydroxide in a neutralization reaction? 2 NaOH + H 2 SO 4  2 H 2 O(s) + Na 2 SO 4 120. g NaOH x x g NaOH mol NaOH mol Na 2 SO 4 = 1 mol MM (g) From PT = 6.02 x 10 23 Atoms Molecules Formula units = = Liters of gas at STP 22.4 40.00 1.50 120. g NaOH ??mol Na 2 SO 4 mol Na 2 SO 4 ??mol NaOH USE MOLE RATIO USE MM MM mol NaOH 1 2 1

15 Using States to Determine How to go to Moles -- Solids (Use the Molar Mass) -- Liquids (Use Density and then Molar Mass) -- Gases at STP (Use 22.4 L) -- Gases at non-STP conditions (Use Ideal Gas Law) -- Aqueous (Use Molarity X Volume in Liters)

16 Solution Stoichiometry (CHECKLIST) If 25.00 mL of Hydrochloric Acid with a concentration of 0.1234 M is neutralized by 23.45 mL of Calcium Hydroxide, what is the concentration (Molarity) of the Calcium Hydroxide base? 2 HCl (aq) + Ca(OH) 2 (aq)  2 H 2 O (l) + CaCl 2 (aq).02500 L x mol HCl L M Ca(OH) 2 = 0.06578 -- Go to moles (Tape) -- Mole ratio (Reaction) -- Go away from moles (Tape) 1 mol HCl.02500 L HCl 0.1234 M HCl ??? M Ca(OH) 2 mol Ca(OH) 2 1 2 x x 0.02345 0.1234 L Recall, the Molarity is mol / L To find Molarity, use a scribble with L

17 Solution Stoichiometry (MOLE MAP) If 25.00 mL of Hydrochloric Acid with a concentration of 0.1234 M is neutralized by 23.45 mL of Calcium Hydroxide, what is the concentration (Molarity) of the Calcium Hydroxide base? 2 HCl (aq) + Ca(OH) 2 (aq)  2 H 2 O (l) + CaCl 2 (aq).02500 L x mol HCl M Ca(OH) 2 = 0.06578 1 0.1234.02500 L 0.1234 M HCl ??? M Ca(OH) 2 mol HCl mol Ca(OH) 2 USE MOLE RATIO USE MOLARITY USE SCRIBBLE VOLUME 1 2 L mol Ca(OH) 2 0.02345 x x mol HCl L Recall, the Molarity is mol / L To find Molarity, use a scribble with L

18 Thermochemistry Stoichiometry (CHECKLIST) The thermite reaction, as shown below, has a Δ H of ─ 849.0 KJ. Determine the amount of energy change when 25.00 grams of aluminum react with excess iron (III) oxide. Fe 2 O 3 (s) + 2Al (s)  Al­ 2 O 3 (s) + 2Fe(s) + 849.0 KJ 25.00 g Al x x g Al mol Al KJ = -393 1 mol MM (g) From PT = 6.02 x 10 23 Atoms Molecules Formula units = = Liters of gas at STP 22.4 -- Place the heat in the reaction. -- Go to moles (Tape) -- Mole ratio with heat (Reaction) -- Go away from moles (Tape) exothermic 25.00 g Al ??? KJ 2 KJ -849 1 mol Al MM 26.98

19 Thermochemistry Stoichiometry (MOLE MAP) The thermite reaction, as shown below, has a Δ H of ─ 849.0 KJ. Determine the amount of energy change when 25.00 grams of aluminum react with excess iron (III) oxide. Fe 2 O 3 (s) + 2Al (s)  Al­ 2 O 3 (s) + 2Fe(s) + 849.0 KJ 25.00 g Al x x KJ = -393 1 mol MM (g) From PT = 6.02 x 10 23 Atoms Molecules Formula units = = Liters of gas at STP 22.4 KJ exothermic 25.00 g Al mol Al ??? KJ USE MM of Al USE RATIO 1 -849 26.98 2 MM g Al mol Al

20 Stoichiometry by Converting the Reaction X 32.00 1 = 160 g NO How many grams of NO is produced if 12 g of O 2 is combined with excess ammonia? 30.01 1 mol NO g O 2 X g NO g O 2 12 g O 2 = ? g NO Step 4) Write the conversion from the reaction Step 3) Convert 4 mol NO to grams NO by multiplying by MM mol O 2 5 mol NO 4 4 NH 3 + 5 O 2  6 H 2 O + 4 NO Step 1) Figure out what is being asked = Step 2) Convert 5 mol O 2 to grams O 2 by multiplying by MM 160 g O 2 = 120.04 g NO Step 5) Write the final proportion and solve 12 g O 2 ??? g NO 120.04 g NO 160 g O 2 = Answer: 9.0 g NO 120.04 mol O 2

21 Mass of Element to Mass of Compound These problem can be done in one step by using a MASS RATIO which you can get from your MOLAR MASS calculation. How many grams of phosphorous are present in 256.8 grams of calcium phosphate? 256.8 51.28 = x g of P g P 61.94 MM of Ca 3 (PO 4 ) 2 Ca = 3 x 40.08 = 120.24 P = 2 x 30.97 = 61.94 O = 8 x 16.00 = + 128 310.18 g/mol g Ca 3 (PO 4 ) 2 g P g Ca 3 (PO 4 ) 2 310.18

22 Acid/Base Stoichiometry (MODIFIED VM=V. M. ) If 25.00 mL of Hydrochloric Acid with a concentration of 0.1234 M is neutralized by 23.45 mL of Calcium Hydroxide, what is the concentration (Molarity) of the Calcium Hydroxide base? 2 HCl (aq) + Ca(OH) 2 (aq)  2 H 2 O (l) + CaCl 2 (aq) x 0.06578 M Ca(OH) 2 = M b = 25.00 mL HCl (# of H + ) V a M a = (# of OH - )V b M b HCl Ca(OH) 2 MbMb 2 OH - 23.45 ml Ca(OH) 2 x x x 0.1234 M HCl 1 H +

23 Video Resources -- Pressure Pressure Tape https://vimeo.com/166356342https://vimeo.com/166356342 But It Works https://vimeo.com/167114263https://vimeo.com/167114263 -- Metric Conversions Metric Tape https://vimeo.com/168331657https://vimeo.com/168331657 Transitive Law https://vimeo.com/168333510https://vimeo.com/168333510 Proportions https://vimeo.com/167111081https://vimeo.com/167111081 --Compound Stoichiometry (Mole Tape) https://vimeo.com/167110420 https://vimeo.com/167110420 Chemical Reactions (Mole Ratio) https://vimeo.com/167110048 https://vimeo.com/167110048

24 Video Resources -- Reaction Stoichiometry Checklist 1 https://vimeo.com/167107604https://vimeo.com/167107604 Mole Map 1 https://vimeo.com/167107216https://vimeo.com/167107216 Checklist 2 https://vimeo.com/167106728https://vimeo.com/167106728 Mole Map 2 https://vimeo.com/167106309https://vimeo.com/167106309 -- Solution Stoichiometry Checklist https://vimeo.com/168332177https://vimeo.com/168332177 Mole Map https://vimeo.com/168332576https://vimeo.com/168332576 -- Thermochemistry Stoichiometry Checklist https://vimeo.com/166991562https://vimeo.com/166991562 Mole Map https://vimeo.com/166988720https://vimeo.com/166988720 -- Stoichiometry by Converting the Reaction https://vimeo.com/166953338https://vimeo.com/166953338 -- Mass of Element to Mass of Compound https://vimeo.com/166950996https://vimeo.com/166950996 -- Acid/Base Stoichiometry Modified VM= VM https://vimeo.com/166946639https://vimeo.com/166946639

Download ppt "Stoichiometry for Strugglers Stoichiometry for Strugglers Website: allenisd.org/Page/49042 by Daniel Haradem & Kathy Vondracek of Allen ISD for CAST2016."

Similar presentations

III. Stoichiometry Stoy – kee – ahm –eh - tree

III. Stoichiometry Stoy – kee – ahm –eh - tree
Stoichiometry Chapter 12.

Stoichiometry Chapter 12.
Chemical Stoichiometry

Chemical Stoichiometry
Unit 4 Lecture 1- Balancing Eqns and the Mole

Unit 4 Lecture 1- Balancing Eqns and the Mole
Unit 5 Moles and Stoichiometry Lesson 2: The Molar Relationships.

Unit 5 Moles and Stoichiometry Lesson 2: The Molar Relationships.
Section 1.3.  Limiting reactant  Excess reactant  Yields:  Theoretical yield  Percentage yield  Experimental yield  Avogadro’s law  Molar volume.

Section 1.3.  Limiting reactant  Excess reactant  Yields:  Theoretical yield  Percentage yield  Experimental yield  Avogadro’s law  Molar volume.
Objectives SWBAT define stoichiometry. SWBAT write a mole ratio relating two substances in a chemical reaction. SWBAT calculate the mass of a reactant.

Objectives SWBAT define stoichiometry. SWBAT write a mole ratio relating two substances in a chemical reaction. SWBAT calculate the mass of a reactant.
Review: Molar Mass of Compounds

Review: Molar Mass of Compounds
Quantitative Relationships (Stoichiometry). Lets take a moment… sit back… relax… and review some previously learned concepts… Lets take a moment… sit.

Quantitative Relationships (Stoichiometry). Lets take a moment… sit back… relax… and review some previously learned concepts… Lets take a moment… sit.
2 Amounts of Substance Learning Objectives: 2.1 A r & M r, Avogadro’s number, and the mole 2.2 Ideal Gas Law 2.3 Empirical and Molecular Formula 2.4 Moles.

2 Amounts of Substance Learning Objectives: 2.1 A r & M r, Avogadro’s number, and the mole 2.2 Ideal Gas Law 2.3 Empirical and Molecular Formula 2.4 Moles.
Student will learn: mole stoichiometry problems mass stoichiometry problems volume stoichiometry problems Student will learn: to calculate amount of reactants.

Student will learn: mole stoichiometry problems mass stoichiometry problems volume stoichiometry problems Student will learn: to calculate amount of reactants.
What quantities are conserved in chemical reactions? grams and atoms.

What quantities are conserved in chemical reactions? grams and atoms.
Quantitative Chemistry

Quantitative Chemistry
2 Amounts of Substance Learning Objectives: 2.1 A r & M r, Avogadro’s number, and the mole 2.2 Ideal Gas Law 2.3 Empirical and Molecular Formula 2.4 Moles.

2 Amounts of Substance Learning Objectives: 2.1 A r & M r, Avogadro’s number, and the mole 2.2 Ideal Gas Law 2.3 Empirical and Molecular Formula 2.4 Moles.
2 pt 3 pt 4 pt 5pt 1 pt 2 pt 3 pt 4 pt 5 pt 1 pt 2pt 3 pt 4pt 5 pt 1pt 2pt 3 pt 4 pt 5 pt 1 pt 2 pt 3 pt 4pt 5 pt 1pt ConversionsGas Laws Molar Mass Equations.

2 pt 3 pt 4 pt 5pt 1 pt 2 pt 3 pt 4 pt 5 pt 1 pt 2pt 3 pt 4pt 5 pt 1pt 2pt 3 pt 4 pt 5 pt 1 pt 2 pt 3 pt 4pt 5 pt 1pt ConversionsGas Laws Molar Mass Equations.
Stoichiometry. Chemical Equations Short hand way to represent chemical reactions H 2 + Cl 2 → HCl Symbols + = reacts with → = produces, yields Δ = adding.

Stoichiometry. Chemical Equations Short hand way to represent chemical reactions H 2 + Cl 2 → HCl Symbols + = reacts with → = produces, yields Δ = adding.
Gas Stoichiometry. Molar Volume of Gases The volume occupied by one mole of a gas at STP (standard temperature and pressure) –Equal to 22.4 L / mol –Can.

Gas Stoichiometry. Molar Volume of Gases The volume occupied by one mole of a gas at STP (standard temperature and pressure) –Equal to 22.4 L / mol –Can.
If you are traveling at 65 mi/h how long will it take to travel 112 km? If your car gets 28 miles per gallon how many liters of gas will it take to travel.

If you are traveling at 65 mi/h how long will it take to travel 112 km? If your car gets 28 miles per gallon how many liters of gas will it take to travel.
Leave space between each step to add more information. 1.Write a balance chemical equation between the acid and the base. Remember it’s a double replacement.

Leave space between each step to add more information. 1.Write a balance chemical equation between the acid and the base. Remember it’s a double replacement.
Stoichiometry Stoichiometry CDO High School. Stoichiometry Consider the chemical equation: 4NH 3 + 5O 2  6H 2 O + 4NO There are several numbers involved.

Stoichiometry Stoichiometry CDO High School. Stoichiometry Consider the chemical equation: 4NH 3 + 5O 2  6H 2 O + 4NO There are several numbers involved.

Similar presentations

Ads by Google