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Visualizing a Chemical Reaction. 2. 2. Na + Cl 2 NaCl. 10. 5. 10. 10. ?. 10. 5. ___ mole Na. ___ mole Cl 2. ___ mole NaCl. Formation of Ammonia. Ratio of eggs to cookies. Proportional Relationships. 2 1/4 c. flour

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visualizing a chemical reaction
Visualizing a Chemical Reaction

2

2

Na + Cl2 NaCl

10

5

10

10

?

10

5

___ mole Na

___ mole Cl2

___ mole NaCl

proportional relationships

Ratio of eggs to cookies

Proportional Relationships

2 1/4 c. flour

1 tsp. baking soda

1 tsp. salt

1 c. butter

3/4 c. sugar

3/4 c. brown sugar

1 tsp vanilla extract

2 eggs

2 c. chocolate chips

Makes 5 dozen cookies.

I have 5 eggs. How many cookies can I make?

Conversion

Factor

5 dozen

5 eggs

2 eggs

150 cookies

= 12.5 dozen cookies

Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

proportional relationships1
Proportional Relationships
  • Stoichiometry
    • mass relationships between substances in a chemical reaction
    • based on the mole ratio
  • Mole Ratio
    • indicated by coefficients in a balanced equation

2 Mg + O2 2 MgO

Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

stoichiometry steps
Stoichiometry Steps

1. Write a balanced equation.

2. Identify known & unknown.

3. Line up conversion factors.

  • Mole ratio - moles  moles
  • Molar mass - moles  grams
  • Molarity - moles  liters soln
  • Molar volume - moles  liters gas
  • Mole ratio - moles  moles

Core step in all stoichiometry problems!!

4. Check answer.

Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

molar volume at stp

StandardTemperature&Pressure

0°C and 1 atm

Molar Volume at STP

1 mol of a gas=22.4 L

at STP

Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

molar volume at stp1
Molar Volume at STP

LITERS

OF GAS

AT STP

Molar Volume

(22.4 L/mol)

MASS

IN

GRAMS

NUMBER

OF

PARTICLES

MOLES

Molar Mass

(g/mol)

6.02  1023

particles/mol

Molarity(mol/L)

LITERS

OF

SOLUTION

Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

stoichiometry problems
Stoichiometry Problems
  • How many moles of KClO3 must decompose in order to produce 9 moles of oxygen gas?

2KClO3 2KCl + 3O2

? mol

9 mol

9 mol O2

2 mol KClO3

3 mol O2

= 6 mol KClO3

Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

slide9

1. 2 Sb + 3 Cl22 SbCl3

2. 2 Mg + O2 2 MgO

3. CaCl2 Ca + Cl2

4. 2 NaClO3 2 NaCl + 3 O2

5. Fe + 2 HCl FeCl2 + H2

6. CuO + H2 Cu + H2O

7. 2 Al + 3 H2SO4 Al2(SO4)3 + 3 H2

slide10

=

2

3

5 mol

x mol

1. 2 Sb + 3 Cl22 SbCl3

2 x = 15

excess

5 mol

7.5 mol

excess

x mol

x mol

x = 7. 5 mol

How many moles of chlorine gas are required to react with 5 moles of antimony?

3 mol Cl2

x mol Cl2 = 5 mol Sb

=

7.5 mol Cl2

2 mol Sb

How many moles of SbCl3 are produced from 5 moles of antimony and excess Cl2?

2 mol SbCl3

x mol SbCl3 = 5 mol Sb

=

5 mol SbCl3

2 mol Sb

How many moles of SbCl3 are produced from 7.5 moles of Cl2 and excess Sb?

2 mol SbCl3

x mol SbCl3 = 7.5 mol Cl2

=

5 mol SbCl3

3 mol Cl2

slide11

2. 2 Mg + O2 2 MgO

10 mol

x L

x mol

How many moles of magnesium oxide are produced from the burning of 10 mol of Mg?

2 mol MgO

x mol MgO = 10 mol Mg

=

10 mol MgO

2 mol Mg

How many liters of oxygen are needed to burn 10 mol of Mg?

Assume 1 mol O2 = 22.4 L

1 mol O2

22.4 L O2

x L O2 = 10 mol Mg

=

5 mol O2

=

112 L O2

2 mol Mg

1 mol O2

1 mol O2

22.4 L O2

x L O2 = 10 mol Mg

=

112 L O2

2 mol Mg

1 mol O2

slide12

3. CaCl2 Ca + Cl2

8 mol

x mol

How many moles of calcium metal are produced

from the decomposition of 8 mol of calcium chloride?

and chlorine gas

1 mol Ca

x mol Ca = 8 mol CaCl2

=

8 mol Ca

1 mol CaCl2

How many moles of calcium metal and chlorine gas are produced

from the decomposition of 8 mol of calcium chloride?

1 mol Cl2

x mol Cl2 = 8 mol CaCl2

=

8 mol Cl2

1 mol CaCl2

ions in aqueous solution

NO31–

NO31–

NO31–

NO31–

Pb2+

Pb2+

Na1+

Na1+

I1–

I1–

Ions in Aqueous Solution

Print Copy of Lab

Pb(NO3)2(s)

+ H2O(l)

Pb(NO3)2(aq)

Pb2+(aq) + 2 NO31–(aq)

add

water

in solution

dissociation:

NaI(s)

+ H2O(l)

NaI(aq)

Na1+(aq) + I1–(aq)

Mix them and get…

Balance to get overall ionic equation…

Cancel spectator ions to get net ionic equation…

slide14

NO31–

NO31–

NO31–

NO31–

Pb2+

Pb2+

Na1+

Na1+

Na1+

Na1+

I1–

I1–

I1–

I1–

+ 2 Na1+(aq) + 2 I1–(aq)

Pb2+(aq) + 2 NO31–(aq)

+ 2 Na1+(aq)

PbI2(s) + 2 NO31–(aq)

Pb2+(aq) + 2 I1–(aq)

PbI2(s)

Solubility

Chart

Mix them and get…

PbI2 + NaNO3

(aq)

(s)

Pb(NO3)2(aq) + 2 NaI(aq)

+ 2 Na1+(aq)

PbI2(s) + 2 NO31–(aq)

solid

in solution

Balance to get overall ionic equation…

Cancel spectator ions to get net ionic equation…

slide15

Zn(NO3)2(aq) + Ba(OH)2(aq)

+ Ba2+(aq)

Zn(OH)2(s) + 2 NO31–(aq)

OH1–

NO31–

Ba2+

Zn2+

NO31–

OH1–

+ Ba2+(aq) + 2OH1–(aq)

Zn2+(aq) + 2 NO31–(aq)

+ Ba2+(aq)

Zn(OH)2(s) + 2 NO31–(aq)

Zn2+(aq) + 2 OH1–(aq)

Zn(OH)2(s)

Solubility

Chart

Mix together Zn(NO3)2(aq) and Ba(OH)2(aq):

Mix them and get…

Ba(NO3)2 and Zn(OH)2

(ppt)

(aq)

Zn(NO3)2(aq)

Ba(OH)2(aq)

Zn2+(aq) + 2 NO31–(aq)

Ba2+(aq) + 2 OH1–(aq)

Balance to get overall ionic equation…

Cancel spectator ions to get net ionic equation…

slide16

?

6

2

3

+

(NH4)3PO4 + Mg(OH)2 Mg3(PO4)2

NH4OH

ammonium phosphate

ammonium hydroxide

magnesium phosphate

magnesium hydroxide

OH1-

NH41+

Now you try…

AlCl3 + Li2CO3  Al2(CO3)3 + LiCl

3

6

2

slide17

Identify the spectator ions and write a net ionic equation when an aqueous solution of aluminum sulfate is mixed with aqueous ammonium hydroxide.

Al3+

Al3+

NH41+

SO42-

OH1-

OH1-

NH41+

SO42-

aluminum

aluminum

sulfate + hydroxide 

ammonium

ammonium

aluminum hydroxide

+

ammonium sulfate

(ppt)

(aq)

Al2(SO4)3(aq) + 6 NH4OH(aq)  2 Al(OH)3 + 3 (NH4)2SO4

2 Al3+(aq) + 3 SO42-(aq) + 6 NH41+(aq) + 6 OH1-(aq)  2 Al(OH)3(ppt) + 6 NH41+(aq) + 3 SO42-(aq)

“spectator ions”

2 Al3+(aq) + 6 OH1-(aq)  2 Al(OH)3(ppt)

Net Ionic Equation

meaning of coefficients
Meaning of Coefficients

2 atoms Na

1 molecule Cl2

2 molecules NaCl

2 Na + Cl2 2 NaCl

2 g sodium

1 g chlorine

2 g sodium chloride

+

=

2 mol sodium

1 mol chlorine

2 mol sodium chloride

(1 mol Cl2) x (71 g/mol)

(2 mol Na) x (23 g/mol)

(2 mol NaCl) x (58.5 g/mol)

117 g

46 g

71 g

117 g

classes of reactions
Classes of Reactions

Chemical reactions

Precipitation

reactions

Oxidation-Reduction

Reactions

Acid-Base

Reactions

Combustion

Reactions

summary of classes of reactions
Summary of Classes of Reactions

Chemical reactions

Precipitation

reactions

Oxidation-Reduction

Reactions

Acid-Base

Reactions

Combustion

Reactions

Synthesis

reactions

(Reactants are

elements.)

Decomposition

reactions

(Products are

elements.)

summary of classes of reactions1
Summary of Classes of Reactions

Chemical reactions

Precipitation

reactions

Oxidation-Reduction

Reactions

Acid-Base

Reactions

Combustion

Reactions

Synthesis

reactions

Decomposition

reactions

ionic bonding formation of magnesium chloride
IONIC BONDING: Formation of Magnesium Chloride

Cl

Cl

Mg

Mg2+

Cl

Cl

Loses 2e- Each gains 1e- One magnesium ion Two chloride ions

Mg2+ Cl1-

[(2+) 2 (1-) = 0]

MgCl2 magnesium chloride

ionic bonding formation of magnesium chloride1
IONIC BONDING: Formation of Magnesium Chloride

Cl

Cl

Mg2+

Mg

Mg2+

Cl

Cl

Loses 2e- Each gains 1e- One magnesium ion Two chloride ions

Mg2+ Cl1-

[(2+) 2 (1-) = 0]

MgCl2 magnesium chloride

resources chemical equations and reactions
Resources -Chemical Equations and Reactions

Worksheet - vocabulary

Worksheet – Balancing Chemical Equations

Worksheet – Chemical Word Equations

Worksheet – Quantitative Relationships in Chem. Eqns.

Worksheet – Chemical Equations (paragraph)

Worksheet – Real Life Chemistry

Worksheet – Balancing Equations (visual)

Worksheet -

Lab – Ions in Solution

Textbook - questions

Outline(general)

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