Starter S-121 1.2 moles of Iron (III) Hydroxide are used in a reaction. How many grams is that?

Starter S-121

1.2 moles of Iron (III) Hydroxide are used in a reaction. How many grams is that?

Stoichiometry

Chapter 12

12.1 The Arithmetic of Equations

Chapter 12

The basis for solving stoichiometry problems is a balanced chemical reaction

A balanced reaction is used to calculate

How much reactant is needed

How much product is produced


)(3)(2)(2 ggg NHHN )(3)(2)(2 23 ggg NHHN

Stoichiometry – the calculation of quantities

This reaction produces

Ammonia which is used

in fertilizers

Balanced reactions are usually

used to calculate grams

of product or reactant


)(3)(2)(2 23 ggg NHHN

Analysis of the reaction

Atoms – 2 atoms of nitrogen combine with 6 atoms of hydrogen – product is 2 nitrogen and 6 hydrogen

Molecules – 1 molecule of nitrogen gas combines with 3 molecules of hydrogen gas to produce 2 molecules of ammonia


)(3)(2)(2 23 ggg NHHN


Most useful

Moles – 1 mole of nitrogen gas reacts with 3 moles of hydrogen gas to produce 2 moles of ammonia


)(3)(2)(2 23 ggg NHHN


Mass is always conserved in a chemical reaction


)(3)(2)(2 23 ggg NHHN


The key

moles can be converted to grams

grams can be converted to moles

Volume – remember that one mole of gas at STP is 22.4L


)(3)(2)(2 23 ggg NHHN


So if we start with 50g of N2, how many moles do we have

How many moles of NH3 would be produced?

For every 1 mole of N2, 2 moles of NH3


)(3)(2)(2 23 ggg NHHN

molg

molgN 8.1

28

150 2

3

2

32 6.31

28.1 molNH

molN

molNHmolN


How many grams of ammonia are produced


)(3)(2)(2 23 ggg NHHN

32

32 6.31

28.1 molNH

molN

molNHmolN

3

3

33 611

176.3 gNH

molNH

gNHmolNH

Starter S-123

A. Balance the following reactions

H2SO4 + KOH K2SO4 + H2O

B. What is the molar mass of the sulfuric acid.

C. If 20.0g of sulfuric acid is used, how many moles is that?

D. How many moles of water would be produced?

12.2 Chemical Calculations

Chapter 12

12.2 Chemical CalculationsMole Ratios

Come from balanced chemical reactions

Conversion factors derived from the coefficients in the balanced reactions

OHSOOSH 2222 OHSOOSH 2222 2232

2

2

2

3

molSO

molO2

2

2

2

molH O

molH S2

2

3

2

molO

molH S


Mole-Mole Calculations

It is possible to convert from one quantity in a balanced reaction to another using mole ratios

For example – if 3.7 moles of sulfur dioxide is produced, how many moles of oxygen were used?


2 2 2 22 3 2 2H S O SO H O

22 2

2

33.7 5.6

2

molOmolSO molO

molSO

Mass-Mass Calculations

Three steps

1. Convert given mass values into mole values.

2. Use a mole ratio to convert to the moles that the question requests

3. Convert this mole quantity to a mass value


2 2 2 22 3 2 2H S O SO H O

Mass-Mass Calculations

How many grams of Oxygen are needed to produce 30.0g of Sulfur Dioxide?


2 2 2 22 3 2 2H S O SO H O

22 2

2

130.0 0.468

64.06

molSOgSO molSO

gSO

2

2 22

30.468 0.702

2

molOmolSO molO

molSO

2

2 22

31.99880.702 22.5

1

gOmolO gO

molO

Starter S-125


Al(NO3)3 + Na2SO4 Al2(SO4)3+ NaNO3

B. If 50.0g of Aluminum Nitrate reacts, how many grams of Aluminum Sulfate are produced?

Volume-Mass Calculations

Same steps, but volume is converted to moles, or moles to volume

Example: If 4.0L of nitrogen monoxide reacts, how many grams of oxygen are used?


( ) 2( ) 2( )g g gNO O NO ( ) 2( ) 2( )2 2g g gNO O NO

14.0 0.18

22.4

molNOLNO molNO

LNO

22

10.18 0.090

2

molOmolNO molO

molNO

22 2

2

31.99980.090 29

1

gOmolO gO

molO

12.3 Limiting Reagent and Percent Yield

Chapter 12

Limiting Reagent – the reactant that runs out first

Amounts of both reactants are given

Example: 80.0g Copper, 25.0g Sulfur

1. Calculate how many moles would each reactant produce

Copper


SCuSCu 2 SCuSCu 22

molCugCu

molCugCu 26.1

55.63

10.80

SmolCu

molCu

SmolCumolCu 2

2 630.02

126.1


Amounts of both reactants are given

Example: 80.0g Copper, 25.0g Sulfur

1. Calculate how many moles would each reactant produce

Sulfur


SCuSCu 2 SCuSCu 22

molSgS

molSgS 780.0

06.32

10.25

SmolCu

molS

SmolCumolS 2

2 780.01

1780.0


From the reactions

Copper would produce 0.630 mol Cu2S

Sulfur would produce 0.780 mol Cu2S

That means copper will run out first – it is the limiting reagent

Sulfur would be the excess reagent


SCuSCu 2 SCuSCu 22

One for you now

1. Balance the reaction

2. If 75.6g C2H4 reacts with 100.8g O2 – what is the limiting reagent?

Oxygen

3. How many moles of water are produced?

2.10 mol H2O

4. How many grams of water are produced?


OHCOOHC 22242 OHCOOHC 22242 223

4242

4242 70.2

05.28

16.75 HmolC

HgC

HmolCHgC

OmolH

HmolC

OmolHHmolC 2

42

242 40.51

270.2

2

2

22 15.3

00.32

18.100 molO

gO

molOgO

OmolHHC 242 40.5

OmolHmolO

OmolHmolO 2

2

22 10.23

215.3

OmolHO 22 10.2

OgHOmolH

OgHOmolH 2

2

22 8.37

1

02.1810.2

Starter S-126


SiO2 + C SiC + CO

B. 35 grams of silicon dioxide reacts with 10.0g of Carbon, how much carbon monoxide is formed?

Reactions rarely produce as much product as is predicted-reactants can be

impure-reactions may not go

to completion-may compete with

smaller “side” reactions

In some reactions as little as 60% yield is considered a good result


Yield – how much product is produced

Theoretical Yield – the value calculated using stoichiometry

Actual Yield – the

amount of product

that actually forms


Percent Yield – a ratio of actual to theoretical yield

This number must be 100% or less

In lab the actual yield is the result you get

On a test, it will be a number that is given to you

The theoretical yield is calculated using limiting reactants


100%ActualYield

PercentYield xTheoreticalYield

Example: What is the theoretical yield of Calcium Oxide if 24.8g of Calcium Carbonate decomposes in the following reaction.

Balanced


3 2CaCO CaO CO

33 3

3

124.8 0.248

100.1

molCaCOgCaCO molCaCO

gCaCO

3

3

10.248 0.248

1

molCaOmolCaCO molCaO

molCaCO

56.10.248 13.9

1

gCaOmolCaO gCaO

molCaO

Example: What is the percent yield if actual yield is 9.6g?


56.10.248 13.9

1

gCaOmolCaO gCaO

molCaO

100%ActualYield


9.6100%

13.9

gCaOPercentYield x

gCaO

69%PercentYield

Example: What is the theoretical yield if 15.0g of nitrogen reacts with 15.0g of hydrogen in the following reaction?

Balance


2 2 3N H NH

22 2

2

115.0 0.536

28.01

molNgN molN

gN

3

2 32

20.536 1.07

1

molNHmolN molNH

molN

2

2 22

115.0 7.44

2.016

molHgH moldH

gH

2 2 33 2N H NH 2 31.07N molNH

32 3

2

27.44 4.96

3

molNHmolH molNH

molH

2 34.96H molNH

33 3

3

17.0311.07 18.2

1

gNHmolNH gNH

molNH

Example: If the actual yield is 10.5g of NH3 what is the percent yield?


33 3

3

17.0311.07 18.2

1

gNHmolNH gNH

molNH

100%ActualYield


3

3

10.5100% 57.7%

18.2

gNHx

gNH

Starter S-130

5.000 grams of Pb(NO3)2 combine with 5.000 grams of KI.

A. Write and balance the reaction (double replacement)

B. Calculate the limiting and excess reagents

C. Calculate how much lead (II) iodide is produced if the percent yield is 60.0%.

A. Balance the following singe replacement reaction

Mg + Fe(OH)3 Fe + Mg(OH)2

B. If 12 g of magnesium reacts with 24 g of Iron (III) Hydroxide, which is limiting?

C. How many moles of Magnesium Hydroxide is formed?

D. How many grams of Magnesium Hydroxide is formed?

E. What is the percent yield if only 5.83 grams of Magnesium Hydroxide actually form?

S-136

A. Balance the following singe replacement reaction

AgNO3 + CaCl2 AgCl + Ca(NO3)2

B. If 15 g of silver nitrate reacts with 17 g of calcium chloride, which is limiting?

C. How many moles of silver chloride are formed?

D. How many grams of silver chloride are formed?

S-137

A. Balance the following double replacement reaction

AgNO3 + CaCl2 AgCl + Ca(NO3)2

B. If 15 g of silver nitrate reacts with 17 g of calcium chloride, which is limiting?

C. How many moles of silver chloride are formed?

D. How many grams of silver chloride are formed?

S-139

Starter S-121 1.2 moles of Iron (III) Hydroxide are used in a reaction. How many grams is that?

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