balancing reactions
DESCRIPTION
Balancing Reactions. --for unbalanced students. Write the reaction. Butane gas(C 4 H 10 ) burns in oxygen gas to form carbon dioxide gas and water vapor. Write the reaction. Butane gas(C 4 H 10 ) burns in oxygen gas to form carbon dioxide gas and water vapor - PowerPoint PPT PresentationTRANSCRIPT
Balancing Reactions
--for unbalanced students
Write the reaction
• Butane gas(C4H10) burns in oxygen gas to form carbon dioxide gas and water vapor
Write the reaction
• Butane gas(C4H10) burns in oxygen gas to form carbon dioxide gas and water vapor
C4H10(g) + O2(g) CO2(g) + H2O(g)
“and”
“forms”
Write the reaction
• Aqueous sodium bromide and chlorine gas form aqueous sodium chloride and liquid bromine
• Silver (I) oxide decomposes to form solid silver and oxygen gas
• Solid phosphorus burns in oxygen gas to form diphosphorus pentoxide.
Writing a reaction• You must include formulas for all
substances.
• You may include states [(s), (l), (g), (aq)]
• You should balance the reaction
Do not include masses—even if they are part of the problem, they are not part of the reaction
Write the reaction
• Aqueous sodium bromide and chlorine gas form aqueous sodium chloride and liquid bromine
• Silver (I) oxide decomposes to form solid silver and oxygen gas
• Solid phosphorus burns in oxygen gas to for diphosphorus pentoxide.
NaBr (aq) + Cl2 (g) NaCl (aq) + Br2(l)
Ag2O(s) Ag (s) + O2(g)
P4 (s) + O2 (g) P2O5 (s)
But…
NaBr (aq) + Cl2 (g) NaCl (aq) + Br2(l)
+ +
But…
NaBr (aq) + Cl2 (g) NaCl (aq) + Br2(l)
+ +
Seems to contradict Dalton’s Atomic Theory
But…
2NaBr(aq) +Cl2 (g)2NaCl(aq) + Br2(l)
+ +
That’s better. It’s balanced!
But…
2NaBr(aq) +Cl2 (g)2NaCl(aq) + Br2(l)
These are coefficients
But…
2NaBr(aq) +Cl2 (g)2NaCl(aq) + Br2(l)
Coefficients indicate moles or
particles of the substance
But…
2NaBr(aq) +Cl2 (g)2NaCl(aq) + Br2(l)
No coefficient? It means there is a 1
(not written in)
Count your atoms:
• H20
• As2O3
• Ca(NO3)2
• 3 Al2(SO4)3
• 2 NH4NO3
The T diagram
• Na Na
• Br Br
• Cl Cl
NaBr(aq) +Cl2 (g)NaCl(aq) + Br2(l)
Never change a good formula!
1) Count all atoms in reactants and products
2) Fix a count by changing a coefficient
3) Repeat steps 1-2 as needed
The T diagram
• Na 1 Na 1
• Br 1 Br 2
• Cl 2 Cl 1
NaBr(aq) +Cl2 (g)NaCl(aq) + Br2(l)
You need to double the bromine on the left.
You need to double the bromine on the left.
• Do not try this at home or anywhere!
NaBr2 Na2Br NaBr2
• Sodium bromide is NaBr
You need to double the bromine on the left.
• Do not try this at home or anywhere!
NaBr2 Na2Br NaBr2
• Sodium bromide is NaBr
• Two Sodium bromide is 2 NaBr
The T diagram
• Na 1 2 Na 1
• Br 1 2 Br 2
• Cl 2 Cl 1
2NaBr(aq) +Cl2 (g)NaCl(aq) + Br2(l)
Recount
The T diagram
• Na 1 2 Na 1 2
• Br 1 2 Br 2
• Cl 2 Cl 1 2
2NaBr(aq) +Cl2 (g)2NaCl(aq) + Br2(l)
Recount
Easy
• __N2 + __O2 __ NO
• __N2 + __ H2 __ NH3
• __S8 + __ F2 __ SF4
• __ NaI + __Br2 __I2 + __ NaBr
If it gets tough:
• Start with elements that only show up twice.
• If one side is even, and the other is odd—double the odd one.
• Leave hydrogen, oxygen and any pure elements to the very end.
Medium
• __NH3+_NaOCl_N2H4+_NaCl+_H2O
• __Ca2C +__ H2O __CH4+__ Ca(OH)2
• __ NH3 + __ O2 __ NO + __ H2O
• ___C4H10O + ___O2___CO2+___H2O
“Am I being punished?”
• (Later)
Types of reactions
Synthesis or composition
A+BAB
Only 1 product!
Decomposition
AB A+B
Only 1 reactant!
Single replacement
A+BC AC+B
or
D+BC BD+CA pure element in
reactants and products!
Double replacement
AB+CD AD+CB
Two ionic compounds switch partners!
Combustion (burning or oxidation)
AB+O2 AOx+BOy
Oxygen is a reactant!
Other Oxidation/Reduction
AB+CD a whole bunch of other
things.
Double replacement
AB+CD AD+BC
If both AD and BC are ionic compounds that dissolve in water—the product never comes together
Two types of double replacement
AB+CD AD+BC
In a neutralization reaction, A=H+, D=OH-,
so AD=water
Write and balance:
• HCl +NaOH• H2SO4 + KOH
• H3PO4 + Al(OH)3
• HClO4 + Ba(OH)2
A neutralization reaction forms a salt and water
Write and balance:
• HCl +NaOH H2O +NaCl
• H2SO4 + KOH K2SO4 + H2O
• H3PO4 + Al(OH)3 AlPO4 + 3H2O
• 2HClO4 + Ba(OH)2 2H2O+ Ba(ClO4)2
A neutralization reaction forms a salt and water
Two types of double replacement
AB+CD AD+BC
In a precipitation reaction, one of the
products won’t dissolve.
For example:
NaCl(aq)+AgNO3(aq) NaNO3(aq)+AgCl(s)
NaNO3 dissolves in water, AgCl doesn’t
AgCl is the precipitate
No precipitate=no reaction
NaNO3(aq)+KCl(aq)
All of these dissolve in water
Before Contents: aqueous
Na+ Cl- K+ NO3-
No precipitate=no reaction
KNO3(aq)+NaCl(aq)
All of these dissolve in water
Before AfterContents: aqueous
Na+ Cl- K+ NO3-
Better yet:
NaCl(aq)+AgNO3(aq) NaNO3(aq)+AgCl(s)
Ag+(aq) + Cl-(aq) AgCl(s)
The sodium and nitrate ions (not changed in the reaction) are spectator ions, omit them
Do you want to save some trouble?NaCl(aq)+AgNO3(aq) NaNO3(aq)+AgCl(s)
is
Ag+(aq) + Cl-(aq) AgCl(s)
This is the net ionic equation
Review Questions:
• Your lab.
• It’s not a question.
Mole conversions review:
• How many moles of Mg are in 36.0 g Mg?• What is the formula mass of Mg3N2?• What is the mass of .49 moles Mg3N2
• Write and balance: Magnesium metal and nitrogen gas form solid magnesium nitride
• In this reaction: how many moles of magnesium nitride can be made from 1.48 mol of magnesium?
The mass to mass problem
g
g
g
1 mol g
1 molmol
mol
The mass to mass problem
g
g
g
1 mol g
1 molmol
mol
Given Value
The mass to mass problem
g
g
g
1 mol g
1 molmol
mol
Formula Mass
The mass to mass problem
g
g
g
1 mol g
1 molmol
mol
Mole ratio (from a
balanced reaction)
The mass to mass problem
g
g
g
1 mol g
1 molmol
mol
Formula Mass
The mass to mass problem
g
g
g
1 mol g
1 molmol
mol
Answer (probably
wrong)
The mass to mass problem
g
g
g
1 mol g
1 molmol
mol
What mass of liquid bromine is formed by the action of chlorine gas
on 12.5 g NaBr?
g
g
g
1 mol g
1 molmol
mol
What mass of liquid bromine is formed by the action of chlorine gas
on 12.5 g NaBr?
g
g
g
1 mol g
1 molmol
mol
2NaBr(aq) +Cl2 (g)2NaCl(aq) + Br2(l)
Step 1
What mass of liquid bromine is formed by the action of chlorine gas
on 12.5 g NaBr?
g
g
g
1 mol g
1 molmol
mol
2NaBr(aq) +Cl2 (g)2NaCl(aq) + Br2(l)
102.89 g/mol 159.81g/mol
Step 2
What mass of liquid bromine is formed by the action of chlorine gas
on 12.5 g NaBr?
12.5 g102.89 g NaBr 2 mol NaBr
159.81g Br2 1 mol NaBr 1 mol Br2
1 mol Br2
9.71 g
Step 3
Name the reactants and products,
1. Fe2O3(s) + CO (g) FeO (s) + CO2 (g)
2. FeO (s) + CO (g) Fe (s) + CO2 (g)
3. C12H22O11(s)+O2(g) CO2 (g) + H2O (g)
4. Fe (s) + O2 (g) Fe2O3 (s)
5. Ca (s) + H2O (l) Ca(OH)2 (aq) + H2 (g)
Balance the equations
1. Fe2O3(s) + CO (g) FeO (s) + CO2 (g)
2. FeO (s) + CO (g) Fe (s) + CO2 (g)
3. C12H22O11(s)+O2(g) CO2 (g) + H2O (g)
4. Fe (s) + O2 (g) Fe2O3 (s)
5. Ca (s) + H2O (l) Ca(OH)2 (aq) + H2 (g)
Calculate the molecular weight of the 1st reactants and 1st products.
1. Fe2O3(s) + CO (g) FeO (s) + CO2 (g)
2. FeO (s) + CO (g) Fe (s) + CO2 (g)
3. C12H22O11(s)+O2(g) CO2 (g) + H2O (g)
4. Fe (s) + O2 (g) Fe2O3 (s)
5. Ca (s) + H2O (l) Ca(OH)2 (aq) + H2 (g)
Starting with 2.0 grams of of the first reactant, determine the mass
of the first product.
1. Fe2O3(s) + CO (g) FeO (s) + CO2 (g)
2. FeO (s) + CO (g) Fe (s) + CO2 (g)
3. C12H22O11(s)+O2(g) CO2 (g) + H2O (g)
4. Fe (s) + O2 (g) Fe2O3 (s)
5. Ca (s) + H2O (l) Ca(OH)2 (aq) + H2 (g)
Percent Yield
• If you are given an actual yield—compare it to the expected yield (from the mass-to-mass problem)
• Yields should be less than 100% (measurement error otherwise)
• %yield = actual/expected x 100%
C12H22O11(s)+O2(g)CO2(g)+H2O (g)
• 25.0 g sucrose, C12H22O11 (FM=342g/mol), is burned, but only 30.0 g CO2 is recovered.
• What is the percent yield?
C12H22O11(s)+O2(g)CO2(g)+H2O (g)
• 25.0 g sucrose, C12H22O11 (FM=342g/mol), is burned, but only 30.0 g CO2 is recovered.
• What is the percent yield?
Do the mass-to-mass problem to find the expected yield.
Divide the actual/expected, convert to a %
Potassium iodate decomposes to form potassium iodide and oxygen gas
• Write the reaction.
Potassium iodate decomposes to form potassium iodide and oxygen gas
• 50 g of KIO3 decomposes and only 35 g KI is recovered.
• What is the percent yield?
C12H22O11(s)+O2(g)CO2(g)+H2O (g)
• 25.0 g sucrose, C12H22O11 (FM=342g/mol), is burned, but only 30.0 g CO2 is recovered.
• What is the percent yield?
Do the mass-to-mass problem to find the expected yield.
Divide the actual/expected, convert to a %
Why are yields less than 100%?
Why are yields less than 100%?
• Incomplete reaction
• Some of your product is not recovered
• A side reaction could use up some reactant
How could a yield be over 100%?
How could a yield be over 100%?
• Contamination in the product
Limiting reactant
• If you are given the amount of two reactants—you will probably run out of one first.
• Which one?
• Do two mass-to-mass problems, the expected yield is the lesser of the two.
• The limiting reactant gives this answer• The excess (XS) reactant is left over. (How
much?)
Do not assume…
…that the one with more mass is in excess—you might need more of it.
…that the one with more moles is in excess—you might need more of it.
Do not assume…
…that the one with more mass is in excess—you might need more of it.
If 150 g nitrogen and 40 g hydrogen make ammonia…
…that the one with more moles is in excess—you might need more of it.
If 5 moles oxygen and 8 moles hydrogen make water…
Do not assume…
…that the one with more mass is in excess—you might need more of it.
If 150 g nitrogen and 40 g hydrogen make ammonia…hydrogen is in excess
…that the one with more moles is in excess—you might need more of it.
If 5 moles oxygen and 8 moles hydrogen make water…oxygen is in excess
Do not assume…
…that the one with more mass is in excess—you might need more of it.
If 150 g nitrogen and 40 g hydrogen make ammonia… 8g hydrogen is left over
…that the one with more moles is in excess—you might need more of it.
If 5 moles oxygen and 8 moles hydrogen make water… 1 mol oxygen is left over
Starting with 2.0 grams of each reactant, identify the limiting reagent
1. Fe2O3(s) + CO (g) FeO (s) + CO2 (g)
Starting with 2.0 grams of each reactant, identify the limiting reagent
1. Fe2O3(s) + CO (g) FeO (s) + CO2 (g)
2. FeO (s) + CO (g) Fe (s) + CO2 (g)
3. C12H22O11(s)+O2(g) CO2 (g) + H2O (g)
4. Fe (s) + O2 (g) Fe2O3 (s)
5. Ca (s) + H2O (l) Ca(OH)2 (aq) + H2 (g)
The mole to mass problem
g
mol g
1 molmol
mol
The mass to mole problem
gmol
g
1 mol
mol
mol
The mass to volume problem (at STP)
g
L
g
1 mol 22.4L1 molmol
mol
Etc.
g
g
g
1 mol g
1 molmol
mol