reaction stoichiometry.notebook

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REACTION STOICHIOMETRY

COMPOSITION STOICHIOMETRY:

The mass relationships of elements in a compound

REACTION STOICHIOMETRY :

the mass relationships between products and reactants

Sep 13­8:55 AM

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Molar Ratios

Consider the following balanced equation:

2H2O + Ca ­­> Ca(OH)2 + H2

The coefficients tell us the ratios of the amount of compounds in the reaction

Where there is no coefficient, we understand 1

May 1­2:01 PM

All types of reaction stoichiometry calculations require knowing the molar ratios.

Molar ratio: a conversion factor that relates the amounts, in moles, of any two substances involved in a chemical reaction.

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2H2O + Ca ­­> Ca(OH)2 + H2

For every 2 molecules or parts or moles of water we get 1 molecule, or part or mole of cacium hydroxide.

comparing them in a ratio: 2:1

When we discuss the amount in moles, this ratio is known as the molar ratio.

May 1­2:04 PM

Molar ratio, cont'd

We get the molar ratio from the balanced equation.the decomposition of aluminum oxide yields aluminum metal and oxygen gas. What are all of the possible molar ratios?

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2Al2O3 ­­> 4Al (s) + 3O2 (g)

Al2O3 : Al

Al2O3 : O2

Al :O2

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All of the problems we deal with can be summed up into 4 types:

The given and unknown quantities are both in moles

(mole to mole calculations)

amnt of given substance amount unknown substance

( in moles) ( in moles)

Type1.

May 24­7:58 AM

Process:

• write a balanced equation with correct formulas

• set up ratio of known to unknown based

on balanced equation mole ratios

• complete the proportion

• answer = # moles

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az

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The given is in moles and

the unknown is a mass expressed in grams

Type 2

Process:

• write a balanced equation with correct formulas

• set up ratio of known to unknown based

on balanced equation mole ratios

• complete the proportion

• answer = # moles• convert unknown from moles to grams

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Example: page 306

PhotosynthesisPlants use energy from the sun, Water and Carbon Dioxide to make glucose.

What mass, in grams, of glucose is produced when 3.0 mol of water react with enough carbon dioxide?

1. Write a balanced equation with correctly written formulas

2. determine the given and unknown

6H2O + 6CO2 C6H12O2 + 6 O2

May 8­1:28 PM

3. determine the molar ratio of known to unknown in the equation

4. set up a proportion between molar ratio and the given amounts in the problem

6H2O + 6CO2 C6H12O6 + 6 O2

Example: page 306

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Type 3

The given is in grams (mass)

The unknown is an amount in molesProcess:

• write a balanced equation with correct formulas

• convert known to moles

• set up ratio of known to unknown based on balanced equation

• complete the proportion

• answer = # moles

#moles = mass (g) ÷ molar mass

ex pg 309

Example: page 309

Manufacture of nitric acid is the catalytic oxidation of ammonia

Given

mass NH3= 824g?

# mole NO =?

# mole H2O=?

4NH3 (g) + 5O2(g) 4NO (g) + 6 H2O (g)

Step 1: Write balanced equation

Step 2: determine given and unknown

Step 3: determine # moles in the given

mass ÷ molar mass = # moles

g ÷ g/mole = mole

824 g ÷ 17.04 g/mole = 48.3568075117371

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4NH3 (g) + 5O2(g) 4NO (g) + 6 H2O (g)

Example: page 309

4molNH3 =4mol NO x

4x = 4(48.36) mol NO

x = 48.4 mol NO

4molNH3 =6mol H2O

48.36 mol NH3

x

4x = 6(48.36) mole H2O

x = 72.5 mol H2O

May 17­11:54 AM

The given is expressed in grams

and the unknown is asked for in grams.

NaCl

Type 4

Cannot form ratios instoichometry using grams.

?

mass SnF2 = ?

Given

mass HF= 30.00g

Sn(s) + 2HF SnF2 (s)+ H2 (g)

calculate the molar mass of compounds in question

HF= 1.01g/mol+ 18.99g/mol = 20.00 g/mol

SnF2 = 118.711g/mol + 38.00 g/mol= 156.711g/mol

How many moles of HF do we have?

30.00g HF ÷ 20.00 g/mol = 1.5 mol HF

mass ÷ molar mass =#mol

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1. Balance equation

TYPE 42. Form molar ratio between known and unknown­

° convert given mass → moles

­Calc. # moles of unknown

convert mole of unknown → grams

May 21­12:01 PM

what is the molar ratio of given compound to

unknown compound in the balanced equation?

2mol HF

1 molSnF2 Next , form a proportion using the actual amount of given and the molar ratio from the balanced equation

2mol HF = 1.5 mol HF

1 molSnF2 x x = 0.75 mol SnF2

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Finally, convert the # of moles to mass

mass = # moles . molar mass

0.75 mol SnF2 x 156.711g/mol =

with significant figures = 118 g SnF2

117.53325 g SnF2

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Type I: mole to mole

Type 2: Type#1 ; then convert result g to molType 3: convert known g → mol + Type I

Type 4: convert known g → mol + Type I; convert result mol to g

Recap:

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1.5 C sugar +2½ C. flour 100 cookies

What if I only have 1 cup of flour?

how many cookies should I get?

2­1/2 C. flour = 1 c flour

100 cookies ?

100 cookies (1) = 2.5 (?)

? = 40 cookies

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is there a way to predict how many

cookies we will get?

Will one ingredient be more important

than the other?

Consider .65 cup of sugar and

1 cup of flour­ which one determines how many cookies?

Suppose we don't have enough of either sugar or flour to make a full recipe­

May 19­1:04 PM

1.5 C sugar + 2½ C. flour 100 cookies

form 2 mole ratios­

one between the product and each of the reactants:

1.5 C sugar = .65 C sugar 2½ C. flour 1 C. flour

100cookies x 100 cookies x

x = 43.33 cookies

What this shows us is that the flour runs out before the sugar­ so the flour limits how much product we get.

(Our limiting reagent)

lets go back to our original "recipe"

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We bake the cookies and get 30

cookies!

What happened?

Quantitative analysis

We won't be calculating a % error, but a % yield.

What percentage of 40 is 30?

actual yield x 100% = percent yield

experimental

30/40 x 100% = 75%

We got 75% of the cookies we expected to get.

So, Theoretically, we should get 40 cookies­

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Type 1: mole to mole equations

In this type of calculation, we are given a reaction,

usually in words, a known amount of a compound or element

and asked to find the amount of product produced,

or which reactant limited the reaction.

so , how does this relate to chemistry?

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Example: ( page 305 of your book)

In a spacecraft, the carbon dioxide exhaled by the astronauts

can be removed by its reaction with lithium hydroxide, LiOH,

according to the following chemical equation:

write a balanced equation: (which is nicely done for us)

CO2 (g) + 2 LiOH(s) Li2CO3(s) + H2O (l)

May 27­10:46 AM

CO2 + H2O C6H12O6+ 6026 66

known

amt of H20=3moles

?Mass of glucose = ?

Mole ratio from balanced equation

6 mole of H2O

1 mole C6H12O6

From this, set up a ratio :

6 mole of H2O = 3mol H2O

1 mole C6H12O6 x

6x = 3 mol C6H12O6

x = .5 mol C6H12O6

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Next: convert the amount (in moles) to the mass of glucose ( in grams)

To do this we need to know the molar mass of glucose

C6H12O6

C: 6 x 12.01 = 72.06g/mole

H: 12 x 1.01 = 12.01g/mole

O: 6 x 16 = 96. g/mole

180 g /mole

Apr 4­1:29 PM

CO2 + H2O C6H12O6+ 6026 6

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Known.489 mol P mass O2 =?

4x =.489x(5mol O2)

x = .59 mol O2

?

4 mol P = .489 mol P

5mol O2 x

mass O2 = .59 x 32 g/mol

molar mass O2 = 32 g/mol

May 28­2:11 PM

known ?

.489mol P mass P402=?

4 mol P = .489 mol P

1 mol P4O10 x

4 x = 1 mol P4O10

x = .25 mole P4O10

= .25 mol (283 g/mol)

= 70.75 g

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Na2CO3+2HNO3 2NaNO3 + CO2 + H2O

given ?

mass Na2CO3 = 100g amt of NaNO3 = ?

molar mass NaNO3 molar mass of Na2CO3

1xNa = 22.99 g/mol 2 x Na = 45.98 g/mole

1x N = l4.01 g/mol 1 x C = 12.01 g/mole

3 x O = 48 g/mol 3 x16 = 48 g/mol

85 g/mol 106 g/mol

2 mol NaNO3 = 1.176 mole NaNO3

1 mol Na2CO3 x

x = 1.176/2 mol Na2CO3

= .588 mol Na2CO3

100 g÷ 85 g/mol = 1.176 mole

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Theoretical Yield:Ideal Calculations

give us an amount of

product we should get.

Actual yieldAmount of product actually

obtained in an experiment.­

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Percentage yield: Actual Yield x 100%

theoretical yield

May 17­10:45 AM

Ideal: 27. 62g

Actual: 23.72g

% yield: 23.72 g x 100%

27. 62 g85.88%

Example:

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REACTION STOICHIOMETRY

Chapter 9

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Practice problems from chapter 3

If one dozen eggs weight 16.0 ounces, how much does one egg weigh?

Apr 3­12:54 PM

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NaCl

NaCl

NaCl

NaClNa

Cl

NaCl

NaCl

NaCl

NaCl

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