Balancing Reactions

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…


+ +

But…


+ +

Seems to contradict Dalton’s Atomic Theory

But…

2NaBr(aq) +Cl2 (g)2NaCl(aq) + Br2(l)

+ +

That’s better. It’s balanced!

But…


These are coefficients

But…


Coefficients indicate moles or

particles of the substance

But…


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.


• Do not try this at home or anywhere!

NaBr2 Na2Br NaBr2

• Sodium bromide is NaBr


• 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


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


g

g

g

1 mol g

1 molmol

mol

Given Value


g

g

g

1 mol g

1 molmol

mol

Formula Mass


g

g

g

1 mol g

1 molmol

mol

Mole ratio (from a

balanced reaction)


g

g

g

1 mol g

1 molmol

mol

Formula Mass


g

g

g

1 mol g

1 molmol

mol

Answer (probably

wrong)


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


on 12.5 g NaBr?

g

g

g

1 mol g

1 molmol

mol


Step 1


on 12.5 g NaBr?

g

g

g

1 mol g

1 molmol

mol


102.89 g/mol 159.81g/mol

Step 2


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)



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.


C12H22O11(s)+O2(g)CO2(g)+H2O (g)



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…


If 150 g nitrogen and 40 g hydrogen make ammonia…


If 5 moles oxygen and 8 moles hydrogen make water…

Do not assume…


If 150 g nitrogen and 40 g hydrogen make ammonia…hydrogen is in excess


If 5 moles oxygen and 8 moles hydrogen make water…oxygen is in excess

Do not assume…


If 150 g nitrogen and 40 g hydrogen make ammonia… 8g hydrogen is left over


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

Balancing Reactions

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