chemical reaction defn...reaction (a special case of double displacement reaction)! • water is one...

Chemistry 11 – Notes on Chemical Reactions

1

Chemical Reaction Defn:

Chemical Reaction: when starting chemical species form different chemicals.

Evidence to indicate that a chemical reaction has occurred:

• Temperature change

• Different coloured materials may be formed

• New phases may be formed (eg. from solid to gas)

Defn:

Chemical Reaction Equation: an equation that shows the chemicals used up and

produced during a chemical reaction

• General form of a chemical equation is: REACTANTS ! PRODUCTS

Demo on Law of Conservation of Mass:

• Lead (II) Nitrate + Potassium Iodide ! Lead (II) Iodide + Potassium Nitrate

Pb(NO3)2 + KI ! PbI2 + KNO3

• Reactants start off as clear , colourless solutions

• Product is yellow in colour

• The mass before and after the reaction of the flask and its contents is the same

Defn:

Law of Conservation of Mass: In a chemical reaction, the total mass of the

reactants is equal to the total mass of the products

Law of Conservation of Atoms: In a chemical reaction, the total number of

atoms before is equal to the total number of

atoms after a chemical reaction.

Checking your understanding: Which of the following(s) is conserved?

1. Element 2. Mass

3. Atoms 4. Volume


2

Balancing Chemical Equations:

• A chemical equation is balanced if there is the same number of each type of atom on each side of the equation.

Example: (NH4)2S + Pb(NO3)2 ! NH4NO3 + PbS Guideline to balancing equations:

1. Put coefficient to balance (big numbers in front of each chemical)

2. Balance entire groups (eg. polyatomic ions: NO3-, NH4

+) first if they are

found on both sides.

3. Balance metal atoms next .

4. Try to avoid balancing H , and O ’s until the end if possible. These atoms

are often found several times throughout an equation.

5. If there are fractions in the equation (eg. ½ O2), multiply the equation by the

whole number (eg. 2) to eliminate the fraction.

6. Always double check your equation after balancing.

Answer: (NH4)2S + Pb(NO3)2 ! 2 NH4NO3 + PbS

Example #2: N2 + H2 ! NH3 Answer: Balance N first N2 + H2 ! 2NH3

Hydrogens in NH3 now have a coefficient in front of it, check on the other side of the equation for H’s and balance them too! N2 + 3H2 ! 2NH3 Subscripts:

• A chemical equation can also show the states in which the reactants and products

exist. The states of chemicals are shown by including the following

symbols in brackets immediately after the chemical formula.

(s) = solid, (l) = liquid, (g) = gas, (aq) = aqueous [“dissolved in water”]


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Example: Solid zinc reacts with a hydrochloric acid solution to form hydrogen gas and dissolved zinc chloride. Zn (s) + 2HCl (aq) ! H2 (g) + ZnCl2 (aq)

• Note: “crystals”, “powder”, and “precipitate” all mean the phase is a solid ! Defn:

• Precipitate: a solid formed when two liquid or aqueous solutions react. Predicting Products

• The key to predicting products is to be able to identify the type of reaction that the reactants will undergo

Example:

1. NO2 ! ?

• There is only one compound on the reactant side. This is a decompose

tion reaction!

• The compound will break down into its individual elements.

Therefore: 2NO2 ! N2 + 2O2

2. Mg + ZnSO4 ! ?

• There is one element and one compound on the reactant side. This is a

single replacement reaction!

• The metals (or non-metals) will switch places. Therefore: Mg + ZnSO4 ! Zn + MgSO4

3. Al + O2 ! ?

• There are two elements by themselves on the reactant side. This is a

synthesis reaction!

• The elements combine to form one compound.

Therefore: 4Al + 3O2 ! 2Al2O3


4

4. HCl + KOH ! ?

• There is an acid and a base on the reactant side. This is a neutralization

reaction (a special case of double displacement reaction)!

• Water is one product, and the other ions combine to form a salt.

Therefore: HCl + KOH ! KCl + H2O

5. C2H2 + O2 ! ?

• There is a hydrocarbon compound and oxygen on the reactant side. This

is a combustion reaction!

• The products will be CO2 and H2O

Therefore: 2C2H2 + 5O2 ! 4CO2 + 2H2O


5

Type of Reaction Definition General Formula Examples What to look for!

Synthesis (S) Combination of two or more

substances to form a compound

A + B ! AB

C + O2 ! CO2

H2 + Cl2 ! 2HCl

Two elements as reactants.

Decomposition (D) Breaking down a molecule into simpler substances AB ! A + B

2KClO3 ! 2KCl + 3O2

2Ag2O ! 4Ag + O2

One compound as a reactant.

Single Replacement (SR)

Replacing one atom in a compound by another atom AX + B ! BX + A

Cu + 2AgNO3 ! Cu(NO3) 2 + 2Ag

Cl2 + 2KI ! I2 + 2KCl

Element and a compound react.

Double Replacement (DR)

Exchange of atoms or groups between two different

compounds AX + BY ! AY + BX Pb(NO3) 2 + 2KI ! PbI2 + 2KNO3

Two compounds react.

Neutralization (N) Special type of double

replacement. Reaction of an acid with a base

HA + BOH ! H2O + BA H2SO4 + 2KOH ! K2SO4 + 2H2O

HCl + NaOH ! NaCl + H2O

Acid (starts with “H”) and Base (end with

“OH”) react

Combustion (C) Reaction of hydrocarbon with

oxygen to produce carbon dioxide and water

CxHy + O2 ! H2O + CO2 CH4 + 2O2 ! CO2 + 2H2O

C5H12 + 8O2 ! 5CO2 + 6H2O

Hydrocarbon and O2


7

Chemical Reactions and Energy:

• Molecules are held together by chemical bonds. • To break a bond, energy has to be added to the bond. (eg. breaking a string

requires energy from you!) • To form a bond, atoms must give off their excess energy

There are two types of reactions:

• Exothermic: gives off heat to its surroundings. (heat exits from the reactants) Eg. CaCl2 (s) + H2O (l) ! Ca2+ (aq) + 2Cl – (aq) [Heat pack demo]

• Endothermic: absorbs heat from its surroundings. (heat enters the reaction) Eg. NH4NO3 (s) + H2O (l) ! NH4

+ (aq) + NO3 – (aq) [Cold pack demo]

• If more energy is used to break the bonds in the reactants than is given off producing the bonds in the products, the reaction is ENDOTHERMIC.

• If less energy is used to break the bonds in the reactants than is given off producing the bonds in the products, the reaction is EXOTHERMIC.


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Chemistry 11 – Acid / Base Titration Notes

At the beginning of the year we learned how to classify chemical reactions. One of the special double replacement reactions that we came across was called a “Neutralization Reaction.” Remembering back, a neutralization reaction involves a reaction between an acid and a base to give us a salt and water.

Acid + Base ! Salt + Water

In general, acids are compounds that begin with an “H” in their formula. (Organic acids ends with “COOH”.) The following is a list of common acids that you will often see in a chemistry lab:

HF Hydrofluoric Acid H2SO4 Sulphuric Acid

HCl Hydrochloric Acid H2SO3 Sulphurous Acid

HBr Hydrobromic Acid HNO3 Nitric Acid

HI Hydroiodic Acid HNO2 Nitrous Acid

H3PO4 Phosphoric Acid CH3COOH Acetic Acid

H2CO3 Carbonic Acid In general, bases are compounds that end with an “OH” in their formula and are commonly known as hydroxides. The following is a list of common bases that you will often see in a chemistry lab:

NaOH Sodium Hydroxide NH4OH Ammonium Hydroxide

Ca(OH)2 Calcium Hydroxide Al(OH)3 Aluminum Hydroxide

In general, a salt is defined as a substance that is neither an acid nor a base. It is an ionic

compound that does not begin with an “H” or end with an “OH.” (Example: NaCl)

Thus, an example of a neutralization reaction that involves HCl and NaOH is as follows:

HCl + NaOH ! NaCl + HOH Acid Base Salt Water


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Titrations Chemists have developed a special process of finding an unknown concentration of a

chemical in a solution based upon the neutralization reaction. This process is known as

the “Titration.”

In a titration, a solution of known concentration is reacted with a solution of an

unknown concentration (but a known volume), until a desired equivalence point is

reached.

The equivalence point is the point at which all of the acid and base have reacted together

in a solution. In other words, the moles of acid present is equal to the moles of base

present in the solution!

Moles of Acid = Moles of Base

Thus, all that is left in the solution after the reaction is salt and water.

The following is a diagram of a typical acid / base titration experiment setup:


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We use a table and stoichiometry to help us figure out the unknown concentration

after we have collected the data from the experiment.

Example: Consider the reaction: H3PO4 + 2KOH ! K2HPO4 + 2H2O If 0.0198 L of H3PO4 with an unknown concentration is reacted with 0.0250 L of 0.500 M KOH to reach the equivalence point, what is the concentration of H3PO4? Table: H3PO4 KOH

Molarity ? 0.500 M

Volume 0.0198 L 0.0250 L

Road Map: Conc KOH ! Moles KOH ! Moles H3PO4 ! Conc H3PO4 Step 1: Calculate the moles of KOH present. Moles KOH = 0.500 M x 0.0250 L = 0.0125 mol KOH Step 2: Calculate the moles of H3PO4 present. Moles H3PO4 = 0.0125 mol KOH x 1 mol H3PO4 = 0.00625 mol H3PO4 2 mol KOH Step 3: Calculate the concentration of H3PO4. Conc H3PO4 = 0.00625 mol H3PO4 = 0.316 M 0.0198 L In a titration experiment, usually more than one volume data is taken to make sure that the

titration is done accurately. How do we approach a question like this? Let’s take a look at

an example.

Example: Consider the reaction: Ca(OH)2 + 2HCl ! CaCl2 + 2H2O A lab technician titrates a 10.0 mL sample of Ca(OH)2 solution with 0.0156 M HCl. The following are the volume data for the titration:

Trial Volume HCl used 1 15.35 mL 2 15.45 mL 3 15.80 mL


11

What is the molarity of the Ca(OH)2 in the solution? Note: If you are given more than one trial for the volume data, choose at least two trials

that have the volume within ± 0.10 mL and take an average of the two volumes.

Average Volume used = (15.35 + 15.45) mL = 15.40 mL HCl 2 Table: HCl Ca(OH)2

Molarity 0.0156 M ?

Volume 15.40 mL 10.0 mL

Road Map: Conc HCl ! Moles HCl ! Moles Ca(OH)2 ! Conc Ca(OH)2 Step 1: Calculate the moles of HCl present. Moles HCl = 0.0156 M x 0.01540 L = 0.0002402 mol Step 2: Calculate the moles of Ca(OH)2 present. Moles Ca(OH)2 = 0.0002402 mol HCl x 1 mol Ca(OH)2 = 0.0001201 mol 2 mol HCl Step 3: Calculate the concentration of Ca(OH)2. Conc Ca(OH)2 = 0.0001201 mol Ca(OH)2 = 0.0120 M 0.0100 L

Excess and Limiting Reagent Study Guide Review: When 15.0g of CH4 is reacted with an excess of Cl2 according to: CH4+Cl2→ CH3Cl+HCl (1) What is the mass of CH3Cl produced?

Data Table: Compound Molar Mass (g/mol) CH4 16.0 Cl2 71.0 CH3Cl 50.5

_________________ g CH3Cl is produced Limiting Reactant Problem: When 15.0g of CH4 is reacted with 10.0g of Cl2 according to the equation (1), what is the mass of CH3Cl produced?

• 15.0g CH4 ! CH4 mol ! CH3Cl mol ! CH3Cl g • 10.0g Cl2 ! Cl2 mol ! CH3Cl mol ! CH3Cl g

Which road map to use??

_____________ g CH3Cl is produced

There is enough CH4 to make ______________g of CH3Cl, but only enough Cl2 to make ______________g of CH3Cl. So, Cl2 sets a limit on the amount of CH3Cl produced. Therefore, Cl2 is called the _________________________. The amount of Cl2 is used up completely and the reaction stops. On the other hand CH4 is called the excess _________________________ because not all of it will be used and there will be some left over. The smallest amount of CH3Cl calculated (7.11g) is known as the _______________________of product or the _________________________. What calculation did we do to find which reactant was the LR?

• Calculate how much a product* could be made based on each reactant, then compared the result. • The reactant that makes the least amount of product is the LR.

*Caution: When a question asks for the identity of LR and there is more than one product in the reaction, pick a product, and stick to it throughout the calculation until you figure out which reactant is the LR.

Quick Practice:

2H2 + O2 ! 2H2O – Calculate how much H2O can be made using 10.0g of H2 and 50.0g of O2 – Which reactant is the limiting reagent (LR) and which is the excess reagent (ER)? – What is the theoretical yield of H2O?

Next question: When 15.0g of CH4 is reacted with 10.0g of Cl2 according to the equation (1), how much of the excess reactant (CH4) is left? Amount of excess left = Starting mass of CH4 – Actual mass of CH4 used = 15.0g CH4 - ? g CH4 Based on the mass of LR given, find the mass of excess reactant actually used

• 10.0 g Cl2 ! Cl2 mol ! CH4 mol ! CH4 g

___________________ g Cl2 is used

Amount of excess left = 15.0g CH4 -_______g CH4 = ________________g CH4

Quick Practice:

2H2 + O2 ! 2H2O – We calculated how much H2O can be made using 10.0g of H2 and 50.0g of O2. – H2 is the ER and O2 is the LR. – How many gram of H2 is left?

Be able to answer:

a) Which reactant is the limiting reactant? b) What is the expected mass of the product? c) How much of the excess reactant is left?

Percentage Yield • Sometimes 100% of the expected amount of products can’t be obtained from a reaction. • Thus, we would want to know the percentage of the actual amount of product obtained when

compared to the expected amount from our stoichiometry calculations.

• We can express this percentage as “Percentage Yield” of the reaction. Example: When 10.0 g of CaCO3 was used, 9.0 g of product was produced. What was the percentage yield of the reaction? Step 1: Calculate expected product Road Map: g CaCO3 ! mol CaCO3 ! mol CaCl2 ! mol CaCO3 ! g CaCO3 Molar Mass of CaCO3: (40.1 g/mol) + (12.0 g/mol) + (3 x 16.0 g/mol) = 100.1 g/mol Calculate: Step 2: Calculate percentage yield % yield = mass obtained x 100% = 9.0 g x 100% = 90% mass expected 10.0 g Why is percentage yield usually less than 100%?

• beaker to another…etc.

•

Human error (reading measurements inaccurately How can percentage yield be more than 100%?

Percentage Yield = Mass of product obtained x 100% Mass of product expected

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Titrations Practice Worksheet

Find the requested quantities in the following problems:

1) If it takes 54 mL of 0.1 M NaOH to neutralize 125 mL of an HCl solution,what is the concentration of the HCl?

2) If it takes 25 mL of 0.05 M HCl to neutralize 345 mL of NaOH solution,what is the concentration of the NaOH solution?

3) If it takes 50 mL of 0.5 M KOH solution to completely neutralize 125 mL ofsulfuric acid solution (H2SO4), what is the concentration of the H2SO4

solution?

4) Can I titrate a solution of unknown concentration with another solution ofunknown concentration and still get a meaningful answer? Explain youranswer in a few sentences.

5) Explain the difference between an endpoint and equivalence point in atitration.

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Solutions to the Titrations Practice Worksheet

For questions 1 and 2, the units for your final answer should be “M”, or “molar”,because you’re trying to find the molarity of the acid or base solution. To solvethese problems, use M1V1 = M2V2.

1) 0.043 M HCl2) 0.0036 M NaOH

For problem 3, you need to divide your final answer by two, because H2SO4 is adiprotic acid, meaning that there are two acidic hydrogens that need to beneutralized during the titration. As a result, it takes twice as much base toneutralize it, making the concentration of the acid appear twice as large as itreally is.

3) 0.1 M H2SO4

4) You cannot do a titration without knowing the molarity of at least one ofthe substances, because you’d then be solving one equation with two unknowns(the unknowns being M1 and M2).

5) Endpoint: When you actually stop doing the titration (usually, this isdetermined by a color change in an indicator or an indication of pH=7.0 on anelectronic pH probe)

Equivalence point: When the solution is exactly neutralized. It’s importantto keep in mind that the equivalence point and the endpoint are not exactly thesame because indicators don’t change color at exactly 7.0000 pH and pH probesaren’t infinitely accurate. Generally, you can measure the effectiveness of atitration by the closeness of the endpoint to the equivalence point.

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Limiting Reactant and Percent Yield Practice Name________________________________________ 1) Consider the following reaction:

NH4NO3 + Na3PO4 Æ (NH4)3PO4 + NaNO3

Which reactant is limiting, assuming we started with 30.0 grams of ammonium nitrate and 50.0 grams of sodium phosphate. What is the mass of each product that can be formed? What mass of the excess reactant(s) is left over?

2) Consider the following reaction:

CaCO3 + FePO4 Æ Ca3(PO4)2 + Fe2(CO3)3

Which reactant is limiting, assuming we start with 100. grams of calcium carbonate and 45.0 grams of iron (III) phosphate. What is the mass of each product that can be formed? What mass of the excess reactant(s) is left over?

2

3) Write the balanced equation for the reaction given below:

CuCl2 + NaNO3 Æ Cu(NO3)2 + NaCl

a) If 15 grams of copper (II) chloride react with 20. grams of sodium nitrate, how much sodium chloride can be formed?

b) What is the name of the limiting reagent? __________________

c) How much of the excess reagent is left over in this reaction?

d) If 11.3 grams of sodium chloride are formed in the reaction, what is the percent yield of this reaction?

3

4) Write the equation for the reaction of iron (III) phosphate with sodium sulfate to make iron (III) sulfate and sodium phosphate.

a) If you perform this reaction with 25 grams of iron (III) phosphate and an excess of sodium sulfate, how many grams of iron (III) sulfate can you make?

b) If 18.5 grams of iron (III) sulfate are actually made when you do this reaction, what is your percent yield? c) Is the answer from problem b) reasonable? Explain.

d) If you do this reaction with 15 grams of sodium sulfate and get a 65.0% yield, how many grams of sodium phosphate will you make?

4

5. Write the balanced equation for the reaction given below:

C2H6 + O2 Æ CO2 + H2O a) If 16.4 L of C2H6 reacts with 0.980 mol of O2 how many liters of carbon dioxide gas will be produced?

b) How many oxygen atoms will be in this sample of carbon dioxide? c) How many moles of the excess reactant will be left over? d) How many grams of the excess reactant will be left over?

e) How many of each atom that makes up the excess reactant will be left over?

6) Choose three problems from the Ebbing textbook from pp 119-120 on percent and

theoretical yield, complete them and staple them to this page

5

Answer Sheet

1) C

onsider the following reaction:

3 NH4 NO

3 + Na3 PO

4 Æ (NH

4 )3 PO4 + 3 NaNO

3

Answ

er the questions above, assuming w

e started with 30 gram

s of amm

onium

nitrate and 50 grams of sodium

phosphate.

am

monium

nitrate is limiting

18.6 gram

s of amm

onium phosphate, 31.9 gram

s of sodium nitrate

29.5 gram

s of sodium phosphate

2) C

onsider the following reaction:

3 CaCO3 + 2 FePO

4 Æ Ca

3 (PO4 )2 + Fe

2 (CO3 )3

A

nswer the questions at the top of this sheet, assum

ing we start w

ith 100 grams

of calcium carbonate and 45 gram

s of iron (II) phosphate.

iron (III) phosphate is lim

iting

46.3 grams of calcium

phosphate, 43.8 grams of iron (III) carbonate

54.0 gram

s of calcium carbonate

3) W

rite the balanced equation for the reaction given below:

CuCl2 + 2 NaNO

3 Æ Cu(NO

3 )2 + 2 NaCl

a) If 15 gram

s of copper (II) chloride react with 20 gram

s of sodium nitrate,

how m

uch sodium chloride can be form

ed?

To solve this problem determ

ine how m

uch sodium chloride can be m

ade from

each of the reagents by themselves. W

hen you work out how

much

sodium chloride can be m

ade with 15 gram

s of copper (II) chloride, you find that 13 gram

s will be form

ed. When starting w

ith 20 grams of sodium

nitrate, 14 gram

s will be form

ed. Since 13.0 grams is the sm

aller number,

that’s our answer. Please note that rounding differences m

ay cause your answ

ers to be slightly different, so if they are, don’t panic.

b) W

hat is the limiting reagent for the reaction in a)? copper (II) chloride

6

c) H

ow m

uch of the excess reagent is left over in this reaction?

excess reagent remaining = 20 gram

s – 19 grams (13.0 / 13.6)

= 1 grams

d)

If 11.3 grams of sodium

chloride are formed in the reaction described in

problem a), w

hat is the percent yield of this reaction?

11.3/13.0 x 100% = 86.9%

4)

Write the equation for the reaction of iron (III) phosphate w

ith sodium sulfate to

make iron (III) sulfate and sodium

phosphate.

2 FePO4 + 3 Na

2 SO4 Æ

Fe2 (SO

4 )3 + 2 Na3 PO

4

a) If I perform

this reaction with 25 gram

s of iron (III) phosphate and an excess of sodium

sulfate, how m

any grams of iron (III) sulfate can I m

ake?

17.2 grams

b)

If 18.5 grams of iron (III) sulfate are actually m

ade when I do this reaction,

what is m

y percent yield?

(18.5 / 17.2) x 100% = 108%

c) Is the answ

er from problem

b) reasonable? Explain.

No. Any yield over 100%

is a violation of the Law of conservation of m

ass.

d) If I do this reaction w

ith 15 grams of sodium

sulfate and get a 65.0% yield,

how m

any grams of sodium

phosphate will I m

ake?

According to the stoichiometry, the theoretical yield is 11.5 gram

s. M

ultiplying this by 0.650, you get 7.48 grams.

7 5)

2C2 H

6 + 7O2 Æ

4CO2 + 6H

2 O

a) If 16.4 L of C

2 H6 reacts w

ith 0.980 mol of O

2 how m

any liters of carbon dioxide gas w

ill be produced?

12.5 L CO2

b) H

ow m

any oxygen atoms w

ill be in this sample of carbon dioxide?

6.74E

23 oxygen atoms

c) H

ow m

any moles of the excess reactant w

ill be left over?

0.453 mol C

2 H6 excess

d) H

ow m

any grams of the excess reactant w

ill be left over?

13.6 g C2 H

6 excess

e) How

many of each atom

that makes up the excess reactant w

ill be left over?

5.45E23 carbon atom

s left 1.64E

24 hydrogen atoms left

chemical reaction defn...reaction (a special case of double displacement reaction)! • water is one...

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