unit 8

Stoichiometry

Chapter 12

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Our Plan:Begin LabDaily Challenge – Smore’s StoichiometryNotes – Stoichiometry ConversionsWorksheet #1 & Worksheet #2Wrap Up - Pyramid

Homework (Write in Planner):WS#1 & WS#2 Due Monday

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Follow the procedure for Day 1 and begin recording observations

Use the internet to answer the Pre-Lab Questions

Stoy-key-ahm-a-tree

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

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

Using stoichiometry, you can determine the quantities of reactants & products in a reaction from the BALANCED equation.

Like the Recipe!

Do show the number of moles of each substance involved in the rxn.

Do Not indicate the actual number of grams of the substance.

N2 + 3H2 --> 2NH3

Atoms NAtoms N

Atoms HAtoms H

22

66

On both the reactant and product side!

N2 + 3H2 --> 2NH3

Molecule NMolecule N22

Molecules HMolecules H22

Molecules NHMolecules NH33

11

33

22

N2 + 3H2 --> 2NH3

Mole NMole N22

Moles HMoles H22

Moles NHMoles NH33

11

33

22

N2 + 3H2 --> 2NH3

Grams NGrams N22

Grams HGrams H22

Grams NHGrams NH33

2828

66

3434

Which were conserved?

Atoms and Mass, but NOT Moles and Molecules!

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

1 mole N1 mole N22

3 moles H3 moles H22

3 moles H3 moles H22

2 moles NH2 moles NH33

2 moles NH2 moles NH33

1 mole N1 mole N22

NN22 + 3H + 3H22 --> 2NH --> 2NH33

Molecules

Mass

Volume

Moles Given

Moles Unknown

Molecules

Mass

Volume

Molar Ratio

Equation

6.022 x 1023 6.022 x 1023

Molar Mass

PT

Molar Mass

PT

22.4 L/mole

22.4 L/mole

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

Molar Ratio Molar Ratio

EquationEquation

MolesMolesGivenGiven

Moles Moles UnknownUnknown

H2 + I2 --> 2HIIf 4 moles of H2 react, how many moles of HI will form?

4 moles H4 moles H22 XX2 moles HI2 moles HI

1 mole H1 mole H22

== 8 moles HI8 moles HI

HH22 + I + I22 --> 2HI 2HI

If 0.8 moles of HI form, how many moles of I2 were used in the rxn?

0.8 moles HI0.8 moles HI XX2 moles HI2 moles HI

1 mole I1 mole I22==0.4 moles I0.4 moles I22

HH22 + I + I22 --> 2HI 2HI

2H2 + O2 --> 2H2O1.If 1.3 moles of H2O form, how many moles of O2 were used in the rxn? (0.65 moles)

2.If 0.21 moles of H2 react, how many moles of H2O will form? (0.21 moles)

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

MolesMolesGivenGiven

molar molar

ratioratio

equationequationMolesMoles

UnknownUnknown

molar molar

mass mass

PTPTMassMass

UnknownUnknown

C6H12O6 +6O2 --> 6CO2 + 6H2O

What mass of sucrose was used if 3 moles of water formed?

3 moles H3 moles H22OO XX 1 mole C1 mole C66HH1212OO66

6 moles H6 moles H22OO==

90g 90g CC66HH1212OO66

XX 180g C180g C66HH1212OO66

1 mole C1 mole C66HH1212OO66

CC66HH1212OO66 +6O +6O22 --> 6CO 6CO22 + 6H + 6H22OO

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

5.0 moles H5.0 moles H22OO XX 6 moles CO6 moles CO22

6 moles H6 moles H22OO

==

220 g CO220 g CO22

XX 44g CO44g CO22

1 mole CO1 mole CO22

CC66HH1212OO66 +6O +6O22 --> 6CO 6CO22 + 6H + 6H22OO

N2O5 + H2O --> 2HNO31.What mass of water was used to form 2.5 moles of HNO3? (23 g H2O)

2.What mass of HNO3 will form from 12.00 moles of N2O5? (1512 g HNO3)

12.00 moles N12.00 moles N22OO55 XX 2 moles HNO2 moles HNO33

1 mole N1 mole N22OO55

== 1512 g. HNO1512 g. HNO33

XX63g. HNO63g. HNO33

1 mole HNO1 mole HNO33

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

MolesMolesGivenGiven

molar molar

ratioratio

equationequation

MolesMolesUnknownUnknown

molar molar

mass mass

PTPT

MassMassGivenGiven

2NO + 3H2 --> 2NH3 + O2

How many moles of NO were used to form 824g. NH3?

48.5 moles NO48.5 moles NO

XX2 moles NO2 moles NO

2 moles NH2 moles NH33

==

824g. NH824g. NH33 XX17g. NH17g. NH33

1 mole NH1 mole NH33

2NO + 3H2NO + 3H22 --> 2NH 2NH33 + O + O22

29.3 moles H29.3 moles H22

XX3 moles H3 moles H22

1 mole O1 mole O22

==

312g. O312g. O22 XX32g. O32g. O22

1 mole O1 mole O22

2NO + 3H2NO + 3H22 --> 2NH 2NH33 + O + O22

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Purple Safety Shower Mole

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Stoichiometry

Coefficient Molar Mass

Periodic Table Balance Ratio

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Turn in:Worksheet #1 & Worksheet #2

Our Plan:Conversions Table ReviewNotes – Mass to Mass ConversionsWorksheet #3

Homework (Write in Planner):Worksheet #3 due next class

Molecules

Mass

Volume

Moles Given

Moles Unknown

Molecules

Mass

Volume

Molar Ratio

Equation

6.022 x 1023 6.022 x 1023

Molar Mass

PT

Molar Mass

PT

22.4 L/mole

22.4 L/mole

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.

Molecules

Mass

Volume

Moles Given

Moles Unknown

Molecules

Mass

Volume

Molar Ratio

Equation

6.022 x 1023 6.022 x 1023

Molar Mass

PT

Molar Mass

PT

22.4 L/mole

22.4 L/mole

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

MolesMolesGivenGiven

molar molar

ratioratio

equationequation

MolesMolesUnknownUnknown

molar molar

mass mass

PTPT

MassMass GivenGiven

molar molar

mass mass

PTPT

MassMass UnknUnkn

2Na + Cl2Na + Cl22 -->2NaCl -->2NaClFind the mass of table salt produced from 18.6g Na.

47.3g NaCl47.3g NaCl

XX2 moles NaCl2 moles NaCl

2 mole Na2 mole Na

==

18.6g. Na18.6g. Na XX

23g. Na23g. Na

1 mole Na1 mole Na 58.5g NaCl58.5g NaCl

1 mole NaCl1 mole NaClXX

2Na + Cl2Na + Cl22 --> 2NaCl --> 2NaCl

2Al + Fe2Al + Fe22OO33 --> Al --> Al22OO33 + + 2Fe2Fe

Find the mass of aluminum oxide produced from 2.3g of aluminum.

4.3 g Al4.3 g Al22OO33

XX1 mole Al1 mole Al22OO33

2 mole Al2 mole Al

==

2.3g. Al2.3g. Al XX

27g. Al27g. Al

1 mole Al1 mole Al 102g Al102g Al22OO33

1 mole Al1 mole Al22OO33

XX

2Al + Fe2Al + Fe22OO33 --> Al --> Al22OO33 + 2Fe + 2Fe

Methane burns in air by the following reaction: CHCH44 + 2O + 2O22 --> --> COCO22 + 2H + 2H22OO

What mass of water is produced by burning 500. g of methane?

1,130 g H1,130 g H22OO

XX2 mole H2 mole H22OO

1 mole CH1 mole CH44

==

500.g CH500.g CH44XX

16g CH16g CH44

1 mole CH1 mole CH44 18g H18g H22OO

1 mole H1 mole H22OOXX

Complete Worksheet #3 by next class!

Turn in:Worksheet #3

Our Plan:Questions on WS 1-3Stoichiometry Team ReviewQuiz WS 1 – 3Lab Day 2

Homework (Write in Planner):Finish missing worksheets

Molecules

Mass

Volume

Moles Given

Moles Unknown

Molecules

Mass

Volume

Molar Ratio

Equation

6.022 x 1023 6.022 x 1023

Molar Mass

PT

Molar Mass

PT

22.4 L/mole

22.4 L/mole

1.14.782.2,8223.8.4528

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Our Plan:Lab Day 3Turn in Lab Today

Homework (Write in Planner):Lab, if not done

#1 – 1000 mg = 1 g, 1000 mL = 1 L#3 – Convert both the 1g AgNO3 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!

Clicker Review

Turn in:Metal in Water Lab

Our Plan:Relay Race ReviewDaily ChallengeNotes – Stoichiometry of Gases and Molecules

Worksheet #4 Homework (Write in Planner):

WS#4 due next class

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)

2H2 + O2 --> 2H2O

Molecules

Mass

Volume

Moles Given

Moles Unknown

Molecules

Mass

Volume

Molar Ratio

Equation

6.022 x 1023 6.022 x 1023

Molar Mass

PT

Molar Mass

PT

22.4 L/mole

22.4 L/mole

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

MolesMolesGivenGiven

molar molar

ratioratio

equationequation

MolesMolesUnknownUnknown

22.4 L22.4 LVolVol GivenGiven

22.4 L22.4 L VolVol UnknUnkn

Conversion factor: 22.4 L/mole

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

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

3.7 L x 1 mole = 0.17 mole

22.4 L

2 CO + O2 --> 2 CO2

What volume of carbon dioxide will be produced from 2.6 L of oxygen at STP?

Molecules

Mass

Volume

Moles Given

Moles Unknown

Molecules

Mass

Volume

Molar Ratio

Equation

6.022 x 1023 6.022 x 1023

Molar Mass

PT

Molar Mass

PT

22.4 L/mole

22.4 L/mole

2.6 L O2 X 1 mole O2 X

22.4 L O2

2 mole CO2 X

1 mole O2

22.4 L CO2 =

1 mole CO2

5.2 L CO2

2 CO + O2 CO + O22 --> 2 CO --> 2 CO22

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

Molecules

Mass

Volume

Moles Given

Moles Unknown

Molecules

Mass

Volume

Molar Ratio

Equation

6.022 x 1023 6.022 x 1023

Molar Mass

PT

Molar Mass

PT

22.4 L/mole

22.4 L/mole

45.3 g CO X 1 mole CO X

28 g CO

2 mole CO2 X

2 mole CO

22.4 L CO2 =

1 mole CO2

36.2 L CO2

2 CO + O2 CO + O22 --> 2 CO --> 2 CO22

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

19.3 g CO2 X 1 mole CO2 X

44 g CO2

1 mole O2 X

2 mole CO2

22.4 L O2 =

1 mole O2

4.91 L O2

2 CO + O2 CO + O22 --> 2 CO --> 2 CO22

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

In the reaction: 2H2 + O2 --> 2H2O if you have 0.50 grams of hydrogen, how many molecules of water will form?

Molecules

Mass

Volume

Moles Given

Moles Unknown

Molecules

Mass

Volume

Molar Ratio

Equation

6.022 x 1023 6.022 x 1023

Molar Mass

PT

Molar Mass

PT

22.4 L/mole

22.4 L/mole

1.5 x 101.5 x 102323 molecules H molecules H22OO

XX2 mole H2 mole H22OO

2 mole H2 mole H22

==

0.50 g H0.50 g H22XX

2 g H2 g H22

1 mole H1 mole H22

6.022 x 106.022 x 102323 moleculesmolecules

1 mole H1 mole H22OOXX

2H2H22 + O + O22 --> 2H --> 2H22OO

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Complete Worksheet #4 by next class period!

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Molecules

Mass

Volume

Moles Given

Moles Unknown

Molecules

Mass

Volume

Molar Ratio

Equation

6.022 x 1023 6.022 x 1023

Molar Mass

PT

Molar Mass

PT

22.4 L/mole

22.4 L/mole

Pre Lab Review – The balanced equation for the reaction of baking soda and vinegar is:

HC2H3O2 + NaHCO3 --> NaC2H3O2 + CO2 + H2O.

How many liters of carbon dioxide can form from 3.00 g of baking soda (NaHCO3)? Show your work at the top of your lab.

What did you predict would happen?

What did happen? Why?

Smore’s Recipe2 Graham Crackers3 Pieces of Chocolate

1 Marshmallow

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

2.Which ingredient determined this?

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

4.Which ingredient determined this?

Even in abundance, the reactants must still combine in proportion

They follow stoichiometry.

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

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

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

1.Write a balanced equation

2.Convert the mass of reactants to the mass of the product.

3.Whichever produces less product is the limiting reactant.

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

2H2H22O + 2Na --> 2NaOH + HO + 2Na --> 2NaOH + H22

22.2 g NaOH22.2 g NaOH

XX2 mol NaOH2 mol NaOH

2 mol H2 mol H22OO

==

10.0 g 10.0 g HH22OO

XX18g H18g H22OO

1 mol H1 mol H22OO 40g NaOH40g NaOH

1 mol NaOH1 mol NaOHXX

7.83 g NaOH7.83 g NaOH

XX2 mol NaOH2 mol NaOH

2 mol Na2 mol Na

==

4.50 g 4.50 g NaNa XX

23g Na23g Na

1 mol Na1 mol Na 40g NaOH40g NaOH

1 mol NaOH1 mol NaOHXX

Na= L.RNa= L.R..

Identify the L.R when 8.90 g HF are combined with 14.56g SiO2 in the following reaction:

SiO2 + 4HF --> SiF4 + 2H2O

4.0 g H4.0 g H22OO

XX4 mol HF4 mol HF

2 mol H2 mol H22OO

==

8.9 g HF8.9 g HFXX

20g HF20g HF

1 mol HF1 mol HF 18g H18g H22OO

1 mol H1 mol H22OOXX

8.700 g H8.700 g H22OO

XX2 mol H2 mol H22OO

1 mol SiO1 mol SiO22

==

14.50g SiO14.50g SiO22 XX60g SiO60g SiO22

1 mol SiO1 mol SiO22 18g H18g H22OO

1 mol H1 mol H22OOXX

HF= L.R.HF= L.R.

SiOSiO22 + 4HF -->SiF + 4HF -->SiF4 4 + 2H+ 2H22OO

Finish Balloon Races Questions

Worksheet #5 Chemistry Cartoon

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Our Plan:L.R. Review ProblemPercent Yield NotesWorksheet #6

Homework (Write in Planner):Worksheet #6 Due Tuesday/Wednesday

Acrylonitrile, C3H3N, is an important ingredient in the production of various fibers and plastics. Acrylonitrile is produced from the following reaction:

C3H6 + NH3 + O2 --> C3H3N + H2O

If 850 g of C3H6 is mixed with 300. g of NH3 and unlimited O2, to produce 850. g of acrylonitrile, what is the percent yield? You must first balance the equation. 91 %

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

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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?

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Theoretical Yield is based on calculations

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

% yield = Actual yield X 100 Theoretical yield

1.Write equation2.Calculate the mass of the product that should have been formed. (theoretical yield)

3.Plug numbers into the equation

Determine the % yield for the reaction between 2.80 g Al(NO3)3 & excess NaOH if 0.966g Al(OH)3 is recovered.

Write the equation:Al(NOAl(NO33))33 + 3NaOH --> Al(OH) + 3NaOH --> Al(OH)33 + + 3NaNO3NaNO33

Calculate theoretical yieldCalculate theoretical yield

1.03 g Al(OH)1.03 g Al(OH)33

XX1 mol Al(OH)1 mol Al(OH)33

==

2.80 g 2.80 g Al(NOAl(NO33))33

XX213g Al(NO213g Al(NO33))33

1 mol Al(NO1 mol Al(NO33))33

78g 78g Al(OH)Al(OH)33

XX1 mol Al(NO1 mol Al(NO33))33 1 mol 1 mol

Al(OH)Al(OH)33

Al(NOAl(NO33))33 + 3NaOH --> Al(OH) + 3NaOH --> Al(OH)33 + 3NaNO + 3NaNO33

Plug numbers into equation

% yield% yield ==0.966g0.966g1.03g1.03g XX 100100

== 93.8%93.8%

Determine the percent yield for the reaction between 15.0 g N2 & excess H2 if 10.5g NH3 is produced.

18.2 g NH18.2 g NH33

XX1 mol N1 mol N22

2 mol NH2 mol NH33

==

15.0g N15.0g N22 XX28g N28g N22

1 mol N1 mol N22 17g NH17g NH33

1 mol NH1 mol NH33

XX

%yield%yield ==10.5g NH10.5g NH33

18.2 g NH18.2 g NH33

XX 100100 == 57.7%57.7%

You have to find the limiting reactant first.

Calculate amount of product for both reactants & whichever is less, use as theoretical yield.

Complete Worksheet #6 by next class!

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Our Plan:Stoichiometry Lab TestTest Review

Homework (Write in Planner):Test Review due next classUNIT 8 TEST NEXT CLASS!

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Given the following reaction:                           C3H8    +    5O2    ------->     3CO2      +    

4H2O

If you start with 14.8 g of C3H8 and 3.44 g of O2, determine the limiting reagent

Given the following equation: H3PO4  +   3 KOH    ------>  K3PO4   +  3 H2O

If 49.0 g of H3PO4 is reacted with excess KOH, determine the percent yield of K3PO4 if you isolate 49.0 g of K3PO4.

Given the following equation: Al(OH)3  +3 HCl   --> AlCl3 + 3 H2O

If you start with 50.3 g of Al(OH)3 and you isolate 39.5 g of AlCl3, what is the percent yield?

Given the following equation: H2SO4  +  Ba(OH)2   ---->  BaSO4  +  H2O

If 98.0 g of H2SO4 is reacted with excess Ba(OH)2, determine the percent yield of if 213.7 g of BaSO4 is formed.

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