Visualizing a Chemical Reaction

?

Visualizing a Chemical Reaction

Na + Cl2 NaCl

___ mole Cl2 ___ mole NaCl___ mole Na

2

10 5 10

2

10 5 10

Formation of Ammonia

Proportional Relationships

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

3/4 c. brown sugar1 tsp vanilla extract2 eggs2 c. chocolate chipsMakes 5 dozen cookies.

2 1/4 c. flour1 tsp. baking soda1 tsp. salt1 c. butter3/4 c. sugar

5 eggs

2 eggs= 12.5 dozen cookies

Ratio of eggs to cookies

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150 cookies5 dozen

ConversionFactor

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

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Stoichiometry Steps1. 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

Core step in all stoichiometry problems!!

– Mole ratio - moles moles

4. Check answer.Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

1 mol of a gas=22.4 Lat STP

Molar Volume at STP

Standard Temperature & Pressure0°C and 1 atm

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Molar Volume at STP

Molar Mass(g/mol)

6.02 1023

particles/mol

MASSIN

GRAMSMOLES

NUMBEROF

PARTICLES

LITERSOF

SOLUTION

Molar Volume (22.4 L/mol)

LITERSOF GASAT STP

Molarity (mol/L)

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Stoichiometry Problems• How many moles of KClO3 must decompose

in order to produce 9 moles of oxygen gas?

9 mol O2 2 mol KClO3

3 mol O2

= 6 mol KClO3

2KClO3 2KCl + 3O2 ? mol 9 mol

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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

7.5 mol

1. 2 Sb + 3 Cl2 2 SbCl3

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

5 mol x mol

x mol Cl2 = 5 mol Sb2 mol Sb

3 mol Cl2 = 7.5 mol Cl2

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

x mol SbCl3 = 5 mol Sb2 mol Sb

2 mol SbCl3 = 5 mol SbCl3

excess x mol

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

x mol SbCl3 = 7.5 mol Cl23 mol Cl2

2 mol SbCl3 = 5 mol SbCl3

25 mol

3x mol

=

2 x = 15x = 7. 5 mol

excess

2. 2 Mg + O2 2 MgO

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

x mol MgO = 10 mol Mg2 mol Mg

2 mol MgO= 10 mol MgO

x mol10 mol

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

x L O2 = 10 mol Mg2 mol Mg

1 mol O2 = 5 mol O2

Assume 1 mol O2 = 22.4 L

1 mol O2

22.4 L O2 = 112 L O2

x L

x L O2 = 10 mol Mg2 mol Mg

1 mol O2

1 mol O2

22.4 L O2 = 112 L O2

3. CaCl2 Ca + Cl2

How many moles of calcium metal are produced from the decomposition of 8 mol of calcium chloride?

x mol Ca = 8 mol CaCl21 mol CaCl2

1 mol Ca= 8 mol Ca

x mol8 mol

How many moles of calcium metal and chlorine gas are produced from the decomposition of 8 mol of calcium chloride?

x mol Cl2 = 8 mol CaCl21 mol CaCl2

1 mol Cl2 = 8 mol Cl2

and chlorine gas

Pb2+NO3

1–

Na1+ I1–

Ions in Aqueous Solution

Pb(NO3)2(s) Pb(NO3)2(aq)

Pb2+(aq) + 2 NO31–(aq)

addwater

NaI(s)

+ H2O(l)

dissociation:

+ H2O(l)

Na1+(aq) + I1–(aq)

Mix them and get…

Balance to get overall ionic equation…

Cancel spectator ions to get net ionic equation…

Print Copy of Lab

Print Copy of Lab

NaI(aq)

NO31–

Pb2+

NO31–

NO31–

in solution

Na1+ I1–

Mix them and get…

Pb2+(aq) + 2 NO31–(aq) + 2 Na1+(aq) + 2 I1–(aq) PbI2(s) + 2 NO3

1–(aq) + 2 Na1+(aq)

Pb2+(aq) + 2 I1–(aq) PbI2(s)

Pb(NO3)2(aq) + 2 NaI(aq) PbI2(s) + 2 NO31–(aq) + 2 Na1+(aq)

Balance to get overall ionic equation…

Cancel spectator ions to get net ionic equation…

SolubilityChart

solid

in solution

Pb2+NO3

1–

Na1+ I1–

NO31–

Na1+ I1–

Pb2+

NO31–

Na1+

I1–

NO31–

Na1+

I1–

PbI2 + NaNO3(s) (aq)

Ba2+

OH1–

OH1–

NO31–

NO31–

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

Zn2+(aq) + 2 NO31–(aq) Ba2+(aq) + 2 OH1–(aq)

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

Balance to get overall ionic equation…

Zn2+

Zn(NO3)2(aq) + Ba(OH)2(aq) Zn(OH)2(s) + 2 NO31–(aq) + Ba2+(aq)

Zn2+(aq) + 2 NO31–(aq) + Ba2+(aq) + 2OH1–(aq) Zn(OH)2(s) + 2 NO3

1–(aq) + Ba2+(aq)

Mix them and get…

Zn2+(aq) + 2 OH1–(aq) Zn(OH)2(s)

Cancel spectator ions to get net ionic equation…

SolubilityChart

Ba(NO3)2 and Zn(OH)2(aq) (ppt)

(NH4)3PO4 + Mg(OH)2 Mg3(PO4)2 NH4OH?

ammonium phosphate magnesium hydroxide magnesium phosphate ammonium hydroxide

NH41+ OH1-

+ 62 3

Now you try…

AlCl3 + Li2CO3 Al2(CO3)3 + LiCl32 6

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

2 Al3+(aq) + 3 SO42-(aq) + 6 NH4

1+(aq) + 6 OH1-(aq) 2 Al(OH)3(ppt) + 6 NH41+(aq) + 3 SO4

2-(aq)

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

Net Ionic Equation

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

sulfate + hydroxide aluminum ammoniumaluminum ammonium aluminum hydroxide ammonium sulfate+Al3+ SO4

2- NH41+ OH1- Al3+ SO4

2-NH41+OH1-

(ppt) (aq)

“spectator ions”

Meaning of Coefficients

2 Na + Cl2 2 NaCl2 g sodium 1 g chlorine 2 g sodium chloride+ =

2 mol sodium 1 mol chlorine 2 mol sodium chloride

(2 mol Na) x (23 g/mol) (1 mol Cl2) x (71 g/mol) (2 mol NaCl) x (58.5 g/mol)

46 g 71 g 117 g

117 g

2 atoms Na 1 molecule Cl2 2 molecules NaCl

Classes of Reactions

Chemical reactions

Precipitation reactions

Acid-BaseReactions

Oxidation-ReductionReactions

Combustion Reactions

Summary of Classes of Reactions

Chemical reactions

Precipitation reactions

Acid-BaseReactions

Oxidation-ReductionReactions

Combustion Reactions

Decomposition reactions

(Products are elements.)

Synthesis reactions

(Reactants are elements.)

Summary of Classes of Reactions

Chemical reactions

Precipitation reactions

Acid-BaseReactions

Oxidation-ReductionReactions

Combustion Reactions

Decomposition reactions

Synthesis reactions

IONIC BONDING: Formation of Magnesium Chloride

Mg Mg2+

Cl

Cl

Cl

Cl

Loses 2e- Each gains 1e- One magnesium ion Two chloride ions Mg2+ Cl1-

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

MgCl2 magnesium chloride

Mg2+

IONIC BONDING: Formation of Magnesium Chloride

Mg Mg2+

Cl

Cl

Cl

Cl

Loses 2e- Each gains 1e- One magnesium ion Two chloride ions Mg2+ Cl1-

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

MgCl2 magnesium chloride