03b Stoichiometry 6.pdf

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CHAPTER 3: PART 2

Chemical Equations and Stoichiometry

A chemical change (

a chemical reaction

) converts

one substance into another.

•Chemical reactions involve:

1. Breaking bonds in the reactants. (reactantsare the stuff you start with) 2. Forming new bonds in the products.

(productsare the new stuff you make)

CH4 and O2 CO2 and H2O

A Chemical Equation Uses Specific Notation to

Show a Chemical Reaction

•The reactants ( on the left) and products (on the right) are separated by an arrow that points to the products.

•A plus sign (+) is used to separate individual reactants and products.

2H2(g) + O2(g) 2H2O(l)

reactants

products

Symbols in parentheses to the right of each formula

indicate the state or form in which the substance

exists.

Symbols: (g) gas (l) liquid

(s) solid

(aq) for a substance dissolved in water (aqueous)

liquid

gas gas

2H2(g) + O2(g) 2H2O(l)

Chemical Equations Are Balanced

Law of Conservation of Matter: Atoms are neither created nor destroyed in chemical reactions.

What this law means:

Atoms are not lost or gained. The total #of hydrogen and oxygen atoms is the same on the left and right sides of the equation.

The mass of the reactants must equal the mass of the products.

2H2(g) + O2(g) 2H2O(l)

4.0 g 32.0 g 36.0 g

Coefficients (numbers) are written to the left of the molecular

formula to make the equation obey the conservation Law.

•There must be the same number of each type of atom on the reactant side

and on the product side.

•The coefficient one is never written. When no coefficient appearance before the molecular formula, it is assumed to be one.

•Coefficients must be integers, not fractions or decimals are allowed. •Without coefficients, the equation is said to be unbalanced.

2H2(g) + O2(g) 2H2O(l)

4 H

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Balancing Chemical Equations

Balance a chemical equation by writing coefficients to the left of the molecular formulas until the number of atoms of each element are equal in the reactants and products.

H2(g) + Cl2(g) HCl(g)

H2(g) + Cl2(g) 2HCl(g)

Unbalanced

Balanced 2 H

atoms atoms 2 Cl 1 H & 1 Cl atom

How to Balance a Chemical Equation

Write a balanced equation for the reaction of propane (C3H8) with oxygen (O2) to form carbon dioxide (CO2) and water (H2O).

Write the equation with the correct formulas.

C3H8 + O2 CO2 + H2O

•The subscriptsin a formula can NEVER be changed to balance an equation, because changing a subscript changes the identity of a compound.

Example

Step [1]

How to Balance a Chemical Equation (

CHO

Method)

Balance the equation with coefficients one element at a time. Balance the C’s first:

Balance the H’s next:

Balance the O’s last: Step [2]

How to Balance a Chemical Equation (

CHO

Method)

Check to make sure that the smallest set of whole numbers is used. Step [3]

Balancing Chemical Equations By Inspection

NaBr(

aq

) + Cl

₂

(

aq

)  NaCl(

aq

) + Br

₂

(

aq

)

2

NaBr(

aq

) + Cl

₂

(

aq

) 

2

NaCl(

aq

) + Br

₂

(

aq

)

H

2

O

2

(

aq

) + H

2

S(

aq

)  H

2

O(

l

) + S(

s

)

H

2

O

2

(

aq

) + H

2

S(

aq

) 

2

H

2

O(

l

) + S(

s

)

Balancing Chemical Equations:

Leave the

Simplest Molecules

for

Last



Unbalanced

H

2

S(

g

) +

O

2

(

g

)  SO

2

(

g

) + H

2

O(

l

)



Balanced

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Balancing Chemical Equations:

Start with

the Most

Complicated

Molecule



Unbalanced

Ag

2

CO

3

(

s

)  Ag(

s

) + CO

2

(

g

) + O

2

(

g

)



Balanced

2 Ag

2

CO

3

 4 Ag(

s

) + 2CO

2

(

g

) + O

2

(

g

)

Balancing Chemical Equations That Contain

Polyatomic Ions

: the

MINOH

Method

MINOH: “Me know how to balance equations!”

Metals Ions(Polyatomic) Nonmetals Oxygen Hydrogen



Unbalanced

Na

2

CO

3

(

aq

) + Ca(NO

3

)

2

 NaNO

3

(

aq

) + CaCO

3

(

s

)

Balanced

Na

2

CO

3

(

aq

) + Ca(NO

3

)

2

(

aq

) 

2

NaNO

3

(

aq

) + CaCO

3

(

s)

Balancing Chemical Equations: the CHO Method and

Using Temporary Fractions

C2H6(g) + O2(g) CO2(g) + H2O(l)

Unbalanced

Step 1:

balance C and H C2H6(g) + O2(g) 2CO2(g) + 3H2O(l)

Step 2:

add up O atoms C2H6(g) + O2(g) 2CO2(g) + 3H2O(l)

2 O atoms 4 + 3 = 7 O atoms

Step 3:

temporarily use a fraction to balance the O

C2H6(g) + O7 ₂ 2(g) 2CO2(g) + 3H2O(l)

Step 4:

clear the fraction 2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(l)

Mole Relationships in Chemical Equations

•The coefficientsof a balanced equation indicatehow many molecules of reactants and products are needed and created, respectively, in the reaction.

•A balanced chemical equation alsoindicates the number of moles of reactants and products

2 moles of CO

2 molecule CO 1 molecule O1 mole of O22

2 moles of CO2 2 molecules CO2

Coefficients

are used to form

Mole Ratios

,

which can serve as

Conversion Factors

A molar ratio is a ratio of the moles (from the coefficients of a balanced equation) for any two substances in an equation.

4 Fe(s) + 3 O2(g) 2 Fe2O3(s)

Fe and O2: 4 moles Fe and 3 moles O2 3 moles O2 4 moles Fe

Fe and Fe2O3: 4 moles Fe and 2 moles Fe2O3

2 moles Fe2O3 4 moles Fe

O2 and Fe2O3: 3 moles O2 and 2 moles Fe2O3 2 moles Fe2O3 3 moles O2

Use

the

mole ratios

from the coefficients in the

balanced equation

to convert

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Moles to Moles Calculation

Calculate the number of moles of CO

2

produced when

3.82 moles of CO reacts with an excess of oxygen.

2

2 2 mol CO 2

moles CO produced = 3.82 mol CO = 3.82 mol CO

2 mol CO 

Moles to Moles Calculation

Using the balanced chemical equation, how many

moles of CO are produced from 3.5 moles of C

2

H

6

?

2 C

2

H

6

(

g

) + 5 O

2

(

g

)



4 CO(

g

) + 6 H

2

O(

g

)

7.0 mol CO

Unwanted unit cancels. 3.5 mol C2H6 x 4 mol CO _{2 mol C}

2H6

=

Mole to Mole Conversions

From the Balanced Equation

2 Al

2

O

3



4 Al + 3 O

2

Every time we use 2 moles of Al

2

O

3

we make 3

moles of O

2

2 moles Al

2

O

3

or

3 mole O

2

3 mole O

2

2 moles Al

2

O

3

How many moles of O

2

are produced when 3.34

moles of Al

2

O

3

decompose? (5.01)

Mass in Chemical Reactions

How much do you make?

How much do you need?

•Stoichiometry is the calculation of quantities in chemical reactions based on a mole ratio from the balanced equation. •Remember that the coefficientsin a balanced equation tellus

how many molesof each kind of substance. •We cannot measure moles. What can we do?

•We can convert grams to moles using the molar mass. •Then do the math with the molar ratio using the coefficients from the

balanced equation.

•Then use the molar mass to convert the moles back to grams.

Steps to Follow to Solve a Stoichiometry Problem

1. Make sure that the equation is balanced.

2. Identify known (given) and unknown (what you are trying to find).

3. Convert grams of known tomoles by dividingbythe molar massof the known.

4. Convert moles of known to moles of unknown by multiplying by the molarratioobtained from the coefficients in the balanced equation.

5. Convert moles of unknown tograms of unknown by

multiplyingby the molarmassof the unknown.

grams of known moles of known moles of unknown grams of unknown

mole–mole conversion factor



Avogadro’s number

Mass/Moles/Particles Conversions

Grams of

Substance Substance Moles of or Molecules # of Atoms



molar mass



Avogadro’s number



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How to Convert Moles of Reactant to Grams of Product

Using the balanced equation,

how many grams of O3 (ozone) are formed from 9.0 mol of O2?

3 O2(g) sunlight 2 O3(g)

Moles of

reactant Grams of product

= 290 g O3 mole–mole

conversion factor

molar mass conversion

factor

Mol O2 cancel. Mol O3 cancel. 9.0 mol O2 x 2 mol O_{3 mol O}3 ₂ x 48.0 g O_{1 mol O}₃3

How to Convert Grams of Reactant to Grams of Product

Moles of reactant

Grams of product mole–mole conversion factor

molar mass conversion factor Moles of

product

Grams of reactant molar mass

conversion

factor [1]

[2]

[3]

How to Convert Grams of Reactant to Grams of Product

Ethanol (C2H6O, molar mass 46.1 g/mol) is synthesized by reacting ethylene (C2H4, molar mass 28.1 g/mol) with water. How many grams of ethanol are formed from 14 g of ethylene?

C2H4 + H2O  C2H6O

14 g C2H4 x 28.1 g C1 mol C22HH4 4 x 1 mol C1 mol C22HH6O 4 x 46.1 g C1 mol C22HH66O O Grams of

reactant

molar mass conversion factor

mole–mole conversion factor

molar mass conversion factor

= 23 g C2H6O Grams of _product

Grams C2H4

cancel. Moles Ccancel.2H4 Moles Ccancel.2H6O

Stoichiometry Practice Problems

To make silicon for computer chips, the following reaction is used:

SiCl4 + 2 Mg  2 MgCl2 + Si

a. How many grams of Mg are needed to make 9.3 g of Si? (Mg = 24.3 g/mol and Si = 28.1 g/mol) (? = 16 g)

b. How many grams of SiCl4 are needed to make 9.3 g of Si? (SiCl4 = 170.1 g/mol and Si = 28.1 g/mol) (? = 56 g)

c. How many grams of MgCl2 are produced along with 9.3 g Si?

(MgCl2 = 95.3 g/mol and Si = 28.1 g/mol) (? = 63 g)

Stoichiometry Practice Problems

The US Space Shuttle boosters use this reaction: 3 Al(s) + 3 NH4ClO4 Al2O3 + AlCl3 + 3 NO + 6 H2O

How much Al must be used to react with 65 g of NH4ClO4? (Al = 27.0 g/mol and NH4ClO4= 117.5 g/mol) (? = 15 g)

Stoichiometry Practice Problems

•Sometimes, we do not need to use all of the steps when solving

a stoichiometry problem.

•Suppose we want to determine the mass in grams of NH3 that can form from 2.50 moles of N2?

N2(g) + 3 H2(g)  2 NH3(g)

(Note: in this problem, we do not need to convert grams of given into moles, because the given is already in moles.)

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The

Limiting Reactant is

the Reactant that is

Completely

Used Up

in the Reaction

Determining the Limiting Reactant

1. Determine how much of one reactant is needed to react with a second reactant.

2 H2(g) + O2(g) 2 H2O(l)

chosen to be

“Original Quantity” “Unknown Quantity” chosen to be

There are 4 molecules of H2

in the picture.

Determining the Limiting Reactant

2. Write out the conversion factors that relate the numbers of

moles (or molecules) of reactants.

2 H2(g) + O2(g) 2 H2O(l)

2 molecules H2

1 molecule O2

2 molecules H2

Choose this conversion factor to cancel molecules H2

Determining the Limiting Reactant

3. Calculate the number of moles (molecules) of the second

reactant needed for complete reaction.

2 H2(g) + O2(g) 2 H2O(l)

1 molecule O2

2 molecules H2

4 molecules H2 x = 2 molecules O2

Determining the Limiting Reactant

4. Analyze the two possible outcomes:

If the amount present of the second reactant is less than what is needed, the second reactant is the limiting reactant.

If the amount present of the second reactant is greater than what is needed, the second reactant is in excess.

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Using the balanced equation, determine the limiting reactant when

10.0 g of N2 (28.02 g/mol) reacts with 10.0 g of O2 (32.00 g/mol).

N2(g) + O2(g)  2 NO(g)

1. Convert the number of grams of each reactant into moles using the molar masses.

Determining the Limiting Reactant

N2(g) + O2(g)  2 NO(g)

2. Determine the limiting reactant by choosing N2 as the original quantity and converting to mole O2.

The amount of O2 we started with (0.313 mol) is less than the amount we would need (0.357 mol) so O2 is the limiting reagent.

1 mol O2

1 mol N2

0.357 mol N2 x = 0.357 mol O2 mole–mole

Conversion factor

Consider the reaction between 5 moles of CO and 8 moles of H2

to produce methanol.

How many moles of H2 are necessary in order for all the CO to react?

How many moles of CO are necessary in order for all of the H2 to react?

10 moles of H2 required; 8 moles of H2 available; limiting reactant.

4 moles of CO required; 5 moles of CO available; excess reactant.

2

2 2 mol H 2

moles of H = 5 mol CO_{1 mol CO} = 10 mol H

CO(g) + 2H2(g) → CH3OH(l)

2 2

1 mol CO moles of CO = 8 mol H = 4 mol CO

2 mol H 

Worked Example 8.7

Strategy Convert each of the reactant masses to moles. Use the balanced equation to write the stoichiometric conversion factor and determine which reactant is limiting. Next, determine the number of moles of excess reactant remaining and the number of moles of CO2 produced. Finally, use the appropriate

molar masses to convert moles of excess reactant and moles of CO2 to grams.

Alka-Seltzer tablets contain aspirin, sodium bicarbonate, and citric acid. When they come into contact with water, the sodium bicarbonate (NaHCO3) and citric

acid (H3C6H5O7) react to form carbon dioxide gas, among other products.

3NaHCO3(aq) + H3C6H5O7(aq) → 3CO2(g) + 3H2O(l) + Na3C6H5O7(aq) The formation of CO2 causes the trademark fizzing when the tablets are dropped

into a glass of water. An Alka-Seltzer tablet contains 1.700 g of sodium bicarbonate and 1.000 g citric acid. Determine, for a single tablet dissolved in water, (a) which ingredient is the limiting reactant, (b) what mass of the excess reactant is left over when the reaction is complete, and (c) what mass of CO2

forms.

Worked Example 8.7 (cont.)

Solution

1.700 g NaHCO3×

1.000 g H3C6H5O7×

(a) To determine which reactant is limiting, calculate the amount of citric acid necessary to react completely with 0.02024 mol sodium bicarbonate.

0.02024 mol NaHCO3×

The amount of H3C6H5O7 required to react with 0.02024 mol of NaHCO3 is more

than a tablet contains. Therefore, citric acid is the limiting reactant and sodium bicarbonate is the excess reactant.

1 mol NaHCO3

84.01 g NaHCO3 = 0.02024 mol NaHCO3

1 mol H3C6H5O7

192.12 g

H3C6H5O7

= 0.005205 mol H3C6H5O7

1 mol H3C6H5O7

3 mol NaHCO3 = 0.006745 mol H3C6H5O7

Worked Example 8.7 (cont.)

Solution

(b) To determine the mass of excess reactant (NaHCO3) left over, first calculate

the amount of NaHCO3 that will react:

0.005205 mol H3C6H5O7×

Thus, 0.01562 mole of NaHCO3 will be consumed, leaving 0.00462 mole

unreacted. Convert the unreacted amount to grams as follows:

0.00462 mol NaHCO3×84.01 g NaHCO_{1 mol NaHCO}₃3 = 0.388 g NaHCO3

3 mol NaHCO3

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Worked Example 8.7 (cont.)

Solution

(c) To determine the mass of CO2 produced, first calculate the moles of CO2

produced from the number of moles of limiting reactant (H3C6H5O7) consumed.

0.005205 mol H3C6H5O7×

Convert this amount to grams as follows:

0.01562 mol CO2×

To summarize the results: (a) citric acid is the limiting reactant, (b) 0.388 g sodium bicarbonate remains unreacted, and (c) 0.6874 g carbon dioxide is produced.

44.01 g CO2

1 mol CO2 = 0.6874 g CO2

3 mol CO2

1 mol H3C6H5O7 = 0.01562 mol CO2

Think About It In a problem such as this, it is a good idea to check your work by calculating the amounts of the other products in the reaction. According to the law of conservation of mass, the combined starting mass of the two reactants (1.700 g + 1.000 g = 2.700 g) should equal the sum of the masses of products and leftover excess reactant. In this case, the masses of H2O and Na3C6H5O7 produced are 0.2815 g and 1.343 g,

respectively. The mass of CO2 produced is 0.6874 g [from part (c)] and

the amount of excess NaHCO3 is 0.388 g [from part (b)]. The total,

0.2815 g + 1.343 g + 0.6874 g + 0.388 g, is 2.700 g, identical to the total mass of reactants.

Theoretical, Actual, and Percent Yield

•Theoretical Yield is the maximum amount of product, which is CALCULATED using the balanced equation.

• Because of side reactions and loss of product in handling, you almost never get out 100% of product. The amount isolated from a reaction is called the actual yield.

% yield =

actual yield (g)

theoretical yield (g)

æ

è

ç

ö

ø

÷

´

100

Calculating Percent Yield

Suppose you have prepared cookie dough to make 5 dozen cookies. The phone rings and you answer. While

you talk, a sheet of 12 cookies burns, and you have to throw them out. The rest of the cookies you make are

okay. What is the percent yield of edible cookies?

Theoretical yield: 60 cookies possible Actual yield: 48 cookies to eat

Percent yield: 48 cookies x 100% = 80.% yield 60 cookies

Simple Percent Yield Problem

If the reaction of ethylene with water to form ethanol has a calculated theoretical yield of 23 g of ethanol, what is the percent yield if only 15 g of ethanol are actually formed?

Percent yield = _{theoretical yield}actual yield (g) _{(g) x 100%}

= 15 ₂₃_{g x 100% =}g 65%

03b Stoichiometry 6.pdf

CHAPTER 3: PART 2

CH4 and O2 CO2 and H2O

A Chemical Equation Uses Specific Notation to

exists.

liquid

Chemical Equations Are Balanced

Coefficients (numbers) are written to the left of the molecular

Balancing Chemical Equations

Unbalanced

How to Balance a Chemical Equation

Example

How to Balance a Chemical Equation (

Balancing Chemical Equations By Inspection

Polyatomic Ions

Metals Ions(Polyatomic) Nonmetals Oxygen Hydrogen

Unbalanced

Unbalanced

Mole Relationships in Chemical Equations

2 moles of CO

Coefficients

mole ratios

moles of CO are produced from 3.5 moles of C

7.0 mol CO

Mole to Mole Conversions

Steps to Follow to Solve a Stoichiometry Problem

mole–mole conversion factor

Mass/Moles/Particles Conversions

molar mass

How to Convert Moles of Reactant to Grams of Product

Moles of

How to Convert Grams of Reactant to Grams of Product

14 g C2H4 x 28.1 g C1 mol C22HH4 4 x 1 mol C1 mol C22HH6O 4 x 46.1 g C1 mol C22HH66O O Grams of

Stoichiometry Practice Problems

Stoichiometry Practice Problems

Completely

chosen to be

Determining the Limiting Reactant

2 molecules H2

Determining the Limiting Reactant

1 molecule O2

Determining the Limiting Reactant

Determining the Limiting Reactant

Worked Example 8.7

Worked Example 8.7 (cont.)

Worked Example 8.7 (cont.)

Theoretical, Actual, and Percent Yield

% yield =

actual yield (g)

Simple Percent Yield Problem

2HgO(s) 2Hg(l) + O2(g)

More Difficult Percent Yield Problem

The collected mercury weighed 5.95 g. What was the percentage yield of the reaction?

More Difficult Percent Yield Problem

A sample of HgO weighing 7.22 g was heated. The collected mercury weighed 5.95 g. What was the percentage yield of the reaction?

More Difficult Percent Yield Problem

A sample of HgO weighing 7.22 g was heated. The collected mercury weighed 5.95 g. What was the percentage yield of the reaction?

Worked Example 8.8 (cont.)

Worked Example 8.8 (cont.)