General and Inorganic Chemistry I - Lecture 1nlcd.elte.hu/szalai/pdf/lecture-1-2019.pdfGeneral and Inorganic Chemistry I Lecture 1 IstvÆn Szalai Eötvös University 2019. Outline

General and Inorganic Chemistry ILecture 1

István Szalai

Eötvös University

2019

Outline

1 Introduction

2 Matter and energy

3 Foundations of Chemistry

1 45

Introduction

Informations

Lecture (3+2 hours/week) + laboratory practice (5hours/week)Lecturers:István Szalai (Monday 3 hours)contacts: ELTE Chemistry building 5th �oor room: 5.117email: [email protected]: nlcd.elte.hu/szalai/teaching.htmlNorbert Szoboszlai and Szabolcs Béni (Tuesday 2 hours)

2 45

nlcd.elte.hu/szalai/teaching.html

Informations

Laboratory grade: average result of the Major Tests and theshort testsTheoretical grade: oral exam during the examination period(December-January). The registration for the exam can bemade in the Neptun system. The theoretical exam covers thetopics discussed during the lectures and the laboratorypractices (including calculations).

3 45

Lecture notes and books

Slides: nlcd.elte.hu/szalai/teaching.htmlen.wikibooks.org/wiki/General_ChemistryOpen Textbook Libraryhttps://open.umn.edu/opentextbooks/BookDetail.aspx?bookId=69

4 45

nlcd.elte.hu/szalai/teaching.html

en.wikibooks.org/wiki/General_Chemistry

https://open.umn.edu/opentextbooks/BookDetail.aspx?bookId=69

https://open.umn.edu/opentextbooks/BookDetail.aspx?bookId=69

Chemistry

5 45

Chemistry

Chemistry is the science that describes matter, its properties,the changes it undergoes, and the energy changes thataccompany those processes.

Inorganic chemistryOrganic chemistryPhysical chemistryBiochemistryApplied Chemistry:Analytical chemistry, Pharmaceutical Chemistry, . . .

6 45

Outline of the semester

IntroductionStructure of Atoms and MoleculesChemical Reactions and EquilibriumThermochemistryChemical Kinetics

7 45

Matter and energy

Lecture 1: Introduction and properties ofmatter

Reading:en.wikibooks.org/wiki/General_Chemistry/Properties_of_Matter

8 45

en.wikibooks.org/wiki/General_Chemistry/Properties_of_Matter

en.wikibooks.org/wiki/General_Chemistry/Properties_of_Matter

Matter and Energy

Matter: anything that has both mass and volume.Mass: a measure of an object’s resistance to change in motion(inertia). F = maVolume: a measure of the amount of space occupied by anobject.

States of matter:Gases: They occupy all parts of any vessel in which they arecon�ned. They are capable of in�nite expansion and arecompressed easily. The individual particles are quite farapart.Liquids: The individual particles are con�ned to a givenvolume. A liquid assumes the shape of the container. Theyare very hard to compress.

9 45

Matter and Energy

Matter: anything that has both mass and volume.Mass: a measure of an object’s resistance to change in motion(inertia). F = maVolume: a measure of the amount of space occupied by anobject.

States of matter:Gases: They occupy all parts of any vessel in which they arecon�ned. They are capable of in�nite expansion and arecompressed easily. The individual particles are quite farapart.Liquids: The individual particles are con�ned to a givenvolume. A liquid assumes the shape of the container. Theyare very hard to compress.

9 45

Matter and Energy

States of matter:Solids: They are rigid and have de�nite shapes. Volumes ofsolids do not vary much with changes in temperature. Incrystalline solids the individual particles occupy de�nitepositions in the crystal structure.Plasma: Like a gas, plasma does not have de�nite shape orvolume. Unlike gases, plasmas are electrically conductive,produce magnetic �elds and electric currents, and respondstrongly to electromagnetic forces. Positively charged nucleiswim in a "sea" of freely-moving disassociated electrons

10 45

Matter and Energy

States of matter:

11 45

Matter and Energy

Energy: is a conserved extensive property of a physical system,which cannot be observed directly but can be calculated from itsstate.

Forms of energy: kinetic energy (Ekin = 12mv

2), potentialenergy (e.g. Epot = mgh), heat, electrical energy. . .Law of conservation of matter and energy: There is nochange in the quantity of matter and energy during anychemical or physical change.Mass-energy equivalence is a concept formulated by AlbertEinstein that explains the relationship between mass andenergy.

E = mc2

12 45

Law of conservation of matter and energy

Chemical reaction:2H2 + O2 −−→ 2H2O

Conservation of Matter:2 molH2 + 1 molO2 −−→ 2 molH2OConservation of Mass (mass is not generally conserved):4 gH2 + 32 gO2 −−→ 36 gH2O2H2 + O2 −−→ 2H2O+ 572000 J (energy)m = E/c2 = 572000 J/(3× 108m/s)2 = 6, 35× 10−12 gThe mass associated with chemical amounts of energy is toosmall to measure.

13 45


Chemical reaction:2H2 + O2 −−→ 2H2OConservation of Matter:2 molH2 + 1 molO2 −−→ 2 molH2O

Conservation of Mass (mass is not generally conserved):4 gH2 + 32 gO2 −−→ 36 gH2O2H2 + O2 −−→ 2H2O+ 572000 J (energy)m = E/c2 = 572000 J/(3× 108m/s)2 = 6, 35× 10−12 gThe mass associated with chemical amounts of energy is toosmall to measure.

13 45


Chemical reaction:2H2 + O2 −−→ 2H2OConservation of Matter:2 molH2 + 1 molO2 −−→ 2 molH2OConservation of Mass (mass is not generally conserved):4 gH2 + 32 gO2 −−→ 36 gH2O

2H2 + O2 −−→ 2H2O+ 572000 J (energy)m = E/c2 = 572000 J/(3× 108m/s)2 = 6, 35× 10−12 gThe mass associated with chemical amounts of energy is toosmall to measure.

13 45


Chemical reaction:2H2 + O2 −−→ 2H2OConservation of Matter:2 molH2 + 1 molO2 −−→ 2 molH2OConservation of Mass (mass is not generally conserved):4 gH2 + 32 gO2 −−→ 36 gH2O2H2 + O2 −−→ 2H2O+ 572000 J (energy)

m = E/c2 = 572000 J/(3× 108m/s)2 = 6, 35× 10−12 gThe mass associated with chemical amounts of energy is toosmall to measure.

13 45


Chemical reaction:2H2 + O2 −−→ 2H2OConservation of Matter:2 molH2 + 1 molO2 −−→ 2 molH2OConservation of Mass (mass is not generally conserved):4 gH2 + 32 gO2 −−→ 36 gH2O2H2 + O2 −−→ 2H2O+ 572000 J (energy)m = E/c2 = 572000 J/(3× 108m/s)2 = 6, 35× 10−12 gThe mass associated with chemical amounts of energy is toosmall to measure.

13 45

Physical Properties

A physical change occurs with no change in chemicalcomposition. e.g.: boiling, melting, vaporization. . .Physical properties altered signi�cantly as matter undergoesphysical changes.

Extensive properties: quantity proportional to the quantityof material in the system. mass, volume, total energyIntensive properties: independent of the quantity ofmaterial density, pressure, temperature

14 45

International System of units (SI Units)

length meter mmass kilogram kgtime second selectric current ampere Atemperature Kelvin Kluminous intensity candela cdamount of substance mole molmeter: the lenght equal to the distance traveled by light invacuum in 1/299,792,458 seconds.mol: the amount of substance that contains as many entities(atoms or other particles) as there are atoms in 0.012 kg of pure126C. 1 mole = 6.022× 1023

15 45

Derived Units

area square meter m2

volume cubic meter m3

density kilogram/cubic meter kg/m3

force newton N kg×m/s2pressure pascal Pa N/m2

energy joule J kg×m2/s2electriccharge coulomb C A×selectricpotential di�erence volt V J/(A×s)

16 45

Derived Units

Density:d =

mV

Suppose an object has a mass of 15.0 g and a volume of 10.0 cm3.What is the density?

d =mV =

15.0g10.0 cm3 = 1.50 g/cm3

17 45

Derived Units

Density:d =

mV

Suppose an object has a mass of 15.0 g and a volume of 10.0 cm3.What is the density?

d =mV =

15.0g10.0 cm3 = 1.50 g/cm3

17 45

SI prefixes

1012 tera T 10−1 deci d109 giga G 10−2 centi c106 mega M 10−3 milli m103 kilo k 10−6 micro µ102 hecto h 10−9 nano n101 deka da 10−12 pico p

10−15 femto f10−18 atto a

18 45

SI prefixes

1 kg=1000g (≈ the mass of 1 L water)

1µg=10−6 g=0.000001 g (a typical small sand grain mass is about3µg)1 ng=10−9 g=0.000000001 g (mass of an average human cell)1 nm=10−9m=0.000000001m (a strand of human DNA is 2.5 nmin diameter)

19 45

SI prefixes

1 kg=1000g (≈ the mass of 1 L water)1µg=10−6 g=0.000001 g (a typical small sand grain mass is about3µg)

1 ng=10−9 g=0.000000001 g (mass of an average human cell)1 nm=10−9m=0.000000001m (a strand of human DNA is 2.5 nmin diameter)

19 45

SI prefixes

1 kg=1000g (≈ the mass of 1 L water)1µg=10−6 g=0.000001 g (a typical small sand grain mass is about3µg)1 ng=10−9 g=0.000000001 g (mass of an average human cell)

1 nm=10−9m=0.000000001m (a strand of human DNA is 2.5 nmin diameter)

19 45

SI prefixes

1 kg=1000g (≈ the mass of 1 L water)1µg=10−6 g=0.000001 g (a typical small sand grain mass is about3µg)1 ng=10−9 g=0.000000001 g (mass of an average human cell)1 nm=10−9m=0.000000001m (a strand of human DNA is 2.5 nmin diameter)

19 45

SI prefixes

1 kg=1000g (≈ the mass of 1 L water)1µg=10−6 g=0.000001 g (a typical small sand grain mass is about3µg)1 ng=10−9 g=0.000000001 g (mass of an average human cell)1 nm=10−9m=0.000000001m (a strand of human DNA is 2.5 nmin diameter)

19 45

Measurement and significant figures

If you repeat a particular measurement, you usually do notobtain precisely the same result, because eachmeasurement is subject to experimental error.Precision refers to the closeness of the set of valuesobtained from identical measurements of a quantity.Accuracy refers to the closeness of a single measurement toits true value.Signi�cant �gures are those digits in a measured number (orin the result of a calculation with measured numbers) thatinclude all certain digits plus a �nal digit having someuncertainty.

20 45


All digits are signi�cant except zeros at the beginning of thenumber. Thus, 9.12 cm, 0.912 cm, and 0.00912 cm all containthree signi�cant �gures.Terminal zeros ending at the right of the decimal point aresigni�cant. Each of the following has three signi�cant�gures: 9.00 cm, 9.10 cm, 90.0 cm.

21 45


Multiplication and division. When multiplying or dividingmeasured quantities, give as many signi�cant �gures in theanswer as there are in the measurement with the leastnumber of signi�cant �gures.

Suppose you want to calculate the solubility of a substance (theamount that dissolves in 100 g of water). You �nd that 0.0634gram of the substance dissolves in 25.31 grams of water. Theamount dissolving in 100.0 grams is

100.0g of water× 0.0634g5.31 g of water

Performing it on a pocket calculator you get 0.250493875.

Themeasurement 0.0634 gram has the least number of signi�cant�gures (three). Therefore, you report the answer to threesigni�cant �gures, that is, 0.250 g.

22 45


Multiplication and division. When multiplying or dividingmeasured quantities, give as many signi�cant �gures in theanswer as there are in the measurement with the leastnumber of signi�cant �gures.

Suppose you want to calculate the solubility of a substance (theamount that dissolves in 100 g of water). You �nd that 0.0634gram of the substance dissolves in 25.31 grams of water. Theamount dissolving in 100.0 grams is

100.0g of water× 0.0634g5.31 g of water

Performing it on a pocket calculator you get 0.250493875. Themeasurement 0.0634 gram has the least number of signi�cant�gures (three). Therefore, you report the answer to threesigni�cant �gures, that is, 0.250 g.

22 45


Addition and subtraction. When adding or subtractingmeasured quantities, give the same number of decimalplaces in the answer as there are in the measurement withthe least number of decimal places.

Now consider the addition of 184.2 grams and 2.324 grams. On acalculator, you �nd that the result is 186.524.

But because thequantity 184.2 grams has the least number of decimal places one,whereas 2.324 grams has three, the answer is 186.5 grams.

23 45


Addition and subtraction. When adding or subtractingmeasured quantities, give the same number of decimalplaces in the answer as there are in the measurement withthe least number of decimal places.

Now consider the addition of 184.2 grams and 2.324 grams. On acalculator, you �nd that the result is 186.524. But because thequantity 184.2 grams has the least number of decimal places one,whereas 2.324 grams has three, the answer is 186.5 grams.

23 45


Rounding is the procedure of dropping nonsigni�cant digits in acalculation result and adjusting the last digit reported. Thegeneral procedure is as follows: Look at the leftmost digit to bedropped.

If this digit is 5 or greater, add 1 to the last digit to beretained and drop all digits farther to the right. Thus,rounding 1.2151 to three signi�cant �gures gives 1.22.If this digit is less than 5, simply drop it and all digits fartherto the right. Rounding 1.2143 to three signi�cant �gures gives1.21.

24 45

Chemical properties

They are exhibited by matter as it undergoes changes incomposition:

acidityredox properties. . .

Chemical changes are transformations in which substances areconverted into other substances.3 Br2(l)+ 2 Al(s) −−→ 2 AlBr3(s)

3 Br2(l) + 2 Al(s)→ 2 AlBr3(s)25 45

Classification of matter

Pure Substances:Fixed composition. Cannot be separated into simpler substancesby physical methods.

Elements: cannot be decomposed into simpler substancesby chemical changesCompounds: can be decomposed into simpler substances bychemical changes, always in a de�nite ratio

26 45



Elements: cannot be decomposed into simpler substancesby chemical changes

Compounds: can be decomposed into simpler substances bychemical changes, always in a de�nite ratio

26 45



Elements: cannot be decomposed into simpler substancesby chemical changesCompounds: can be decomposed into simpler substances bychemical changes, always in a de�nite ratio

26 45


Mixtures:Variable composition. Components retain their characteristicproperties. May be separated into pure substances by physicalmethods (e.g. distillation). Mixtures of di�erent compositionsmay have widely di�erent properties.

Homogeneous mixtures: components are indistinguishable;have same composition throughout (e.g. solutions, alloys).Heterogeneous mixtures: components are distinguishable;do not have same composition throughout (e.g. muddy riverwater).

27 45



Homogeneous mixtures: components are indistinguishable;have same composition throughout (e.g. solutions, alloys).

Heterogeneous mixtures: components are distinguishable;do not have same composition throughout (e.g. muddy riverwater).

27 45



Homogeneous mixtures: components are indistinguishable;have same composition throughout (e.g. solutions, alloys).Heterogeneous mixtures: components are distinguishable;do not have same composition throughout (e.g. muddy riverwater).

27 45

Compounds

Law of de�nite proportions: pure compounds always consist ofthe same elements combined in the same proportion by mass.

water 11.1 % hydrogen 88.9 % oxygencarbon dioxide 27.3 % carbon 72.7 % oxygen

28 45

Compounds

Law of Multiple Proportions: If two elements form more than onecompound, in these compounds masses of one element thatcombine with a �xed mass of the other element are in a ratio ofinteger numbers.

N:ON2O 1:0.57NO 1:1.14 0.57:1.14=1:2NO2 1:2.28 0.57:2.28=1:4

29 45

Foundations of Chemistry


Atoms: the smallest particle of an element that maintains itschemical identity.

Structure of atoms:The diameter of an atom is ∼ 10−10 m (0.1 nm).The nucleus contains protons and neutrons. The diameter ofa nucleus is ∼ 10−15 m.

Particle Mass Chargeproton 1.672× 10−27 kg +1.602× 10−19 Cneutron 1.675× 10−27 kg noneelectron 9.109× 10−31 kg −1.602× 10−19 Cm(p+)m(e−) ≈ 1840

30 45


Atoms: the smallest particle of an element that maintains itschemical identity.Structure of atoms:

The diameter of an atom is ∼ 10−10 m (0.1 nm).The nucleus contains protons and neutrons. The diameter ofa nucleus is ∼ 10−15 m.


30 45


Atoms: the smallest particle of an element that maintains itschemical identity.Structure of atoms:

The diameter of an atom is ∼ 10−10 m (0.1 nm).The nucleus contains protons and neutrons. The diameter ofa nucleus is ∼ 10−15 m.


30 45


Atomic number (Z) = number of protons in the nucleus

Mass number (A) = number of protons + number of neutronsSymbol: AZX, 126CIsotopes: atoms for the same elements with di�erentmasses e.g 1

1H, 21H, 31HAtomic mass unit: exactly 1/12 of the mass of an atom of 126C(1 amu = 1.6606× 10−27 kg)On this scale the atomic weight of hydrogen is 1.00794 amu.Relative atomic mass: atomic weight divided by the atomicmass unitRelative atomic weight of hydrogen is 1.00794.

31 45


Atomic number (Z) = number of protons in the nucleusMass number (A) = number of protons + number of neutrons

Symbol: AZX, 126CIsotopes: atoms for the same elements with di�erentmasses e.g 1


31 45


Atomic number (Z) = number of protons in the nucleusMass number (A) = number of protons + number of neutronsSymbol: AZX, 126CIsotopes: atoms for the same elements with di�erentmasses e.g 1

1H, 21H, 31HAtomic mass unit: exactly 1/12 of the mass of an atom of 126C(1 amu = 1.6606× 10−27 kg)On this scale the atomic weight of hydrogen is 1.00794 amu.

Relative atomic mass: atomic weight divided by the atomicmass unitRelative atomic weight of hydrogen is 1.00794.

31 45


Atomic number (Z) = number of protons in the nucleusMass number (A) = number of protons + number of neutronsSymbol: AZX, 126CIsotopes: atoms for the same elements with di�erentmasses e.g 1


31 45


In the molecules two or more atoms are bonded together.

An ion is an atom or group of atoms that carries an electriccharge. Ions that possess a positive charge are calledcations. Those carrying negative charge are called anions.A radical is a group of atoms which have odd number ofelectrons.

32 45


In the molecules two or more atoms are bonded together.An ion is an atom or group of atoms that carries an electriccharge. Ions that possess a positive charge are calledcations. Those carrying negative charge are called anions.

A radical is a group of atoms which have odd number ofelectrons.

32 45


In the molecules two or more atoms are bonded together.An ion is an atom or group of atoms that carries an electriccharge. Ions that possess a positive charge are calledcations. Those carrying negative charge are called anions.A radical is a group of atoms which have odd number ofelectrons.

32 45

Problems

What is the volume of a C atom if its radius is 77 pm?

V =43R

3π

V =43(77× 10

12m)3π

V = 1.91× 10−30 m3

33 45

Problems

What is the volume of a C atom if its radius is 77 pm?

V =43R

3π

V =43(77× 10

12m)3π

V = 1.91× 10−30 m3

33 45

Problems

A beaker weighed 53.10 g. To the beaker was added 5.348 g of ironpellets and 56.1 g of hydrochloric acid. What was the total massof the beaker and the mixture (before reaction)? Express theanswer to the correct number of signi�cant �gures.

Answer: 114.5 g

34 45

Problems

A beaker weighed 53.10 g. To the beaker was added 5.348 g of ironpellets and 56.1 g of hydrochloric acid. What was the total massof the beaker and the mixture (before reaction)? Express theanswer to the correct number of signi�cant �gures.Answer: 114.5 g

34 45

Problems

What is the number of protons, neutrons and electrons in a 23Naatom, in 1g of Na and in 1 mol of Na?

23Na atom: N(p+)=11, N(e+)=11, N(n+)=12

1g Na: N(atoms)= 1g23g/mol6× 10

23 = 2.61× 1022

N(p+)=11× 2.61× 1022=2.87× 1023N(e+)=11× 2.61× 1022=2.87× 1023N(n+)=12× 2.61× 1022=3.13× 1023

1 mol Na: N(atoms)=6× 1023N(p+)=11× 6× 1023=6.6× 1024N(e+)=11× 6× 1023=6.6× 1024N(n+)=12× 6× 1023=7.2× 1024

35 45

Problems


23Na atom: N(p+)=11, N(e+)=11, N(n+)=12


23 = 2.61× 1022

N(p+)=11× 2.61× 1022=2.87× 1023N(e+)=11× 2.61× 1022=2.87× 1023N(n+)=12× 2.61× 1022=3.13× 1023


35 45

Problems


23Na atom: N(p+)=11, N(e+)=11, N(n+)=12


23 = 2.61× 1022

N(p+)=11× 2.61× 1022=2.87× 1023N(e+)=11× 2.61× 1022=2.87× 1023N(n+)=12× 2.61× 1022=3.13× 1023


35 45

Problems


23Na atom: N(p+)=11, N(e+)=11, N(n+)=12


23 = 2.61× 1022

N(p+)=11× 2.61× 1022=2.87× 1023N(e+)=11× 2.61× 1022=2.87× 1023N(n+)=12× 2.61× 1022=3.13× 1023


35 45

Problems

What is the charge of a Na+ ion and 1g of H+?

Na+ ion: Q(Na+) = +1.602× 1019C

1g H+: N = 1g1g/mol6× 10

23mol−1 = 6× 1023

Q = +1.602× 1019C× 6× 1023 = 9.612× 104C

36 45

Problems


Na+ ion: Q(Na+) = +1.602× 1019C

1g H+: N = 1g1g/mol6× 10

23mol−1 = 6× 1023

Q = +1.602× 1019C× 6× 1023 = 9.612× 104C

36 45

Problems


Na+ ion: Q(Na+) = +1.602× 1019C

1g H+: N = 1g1g/mol6× 10

23mol−1 = 6× 1023

Q = +1.602× 1019C× 6× 1023 = 9.612× 104C

36 45

Problems

What is the atomic weight of magnesium according to the abovedata?

Isotope % Abudance Mass (amu)24Mg 78.99 23.98525Mg 10.00 24.98626Mg 11.01 25.983

Atomic weight of magnesium=0.7899× 23.985+ 0.1000× 24.986+ 0.1101× 25.983 = 24.30amu

37 45

Problems




37 45

Problems




37 45

Problems

Cocaine has the following percent composition by mass: 67.30%C, 6.93% H, 21.15% O and 4.62% N. What is the simplest formula ofcocaine?

C 67.3012 = 5.608

H 6.931 = 6.93

O 21.1516 = 1.322

N 4.6214 = 0.33

C 5.6080.33 = 17

H 6.930.33 = 21

O 1.3220.33 = 4

N 0.330.33 = 1

C12H21O4N

38 45

Problems


C 67.3012 = 5.608

H 6.931 = 6.93

O 21.1516 = 1.322

N 4.6214 = 0.33

C 5.6080.33 = 17

H 6.930.33 = 21

O 1.3220.33 = 4

N 0.330.33 = 1

C12H21O4N

38 45

Problems


C 67.3012 = 5.608

H 6.931 = 6.93

O 21.1516 = 1.322

N 4.6214 = 0.33

C 5.6080.33 = 17

H 6.930.33 = 21

O 1.3220.33 = 4

N 0.330.33 = 1

C12H21O4N

38 45

Quiz

Which one of the following is an example of a chemical change?

A. Mixing sand and sugarB. Cutting a piece of paper into two piecesC. Ice melting to waterD. Burning a piece of paper to form carbon dioxide and waterE. Mixing water and orange juice

39 45

Quiz

Which of the following is an example of a solution?

A. WaterB. A combination of red and white chalk dustC. Carbon disul�de (a chemical combination of carbon andsulfur)D. AluminumE. Sugar water

40 45

Quiz

Which of the following are substances?

A. Elements and solutionsB. Elements and compoundsC. Heterogeneous mixtures onlyD. Heterogeneous mixtures and solutionsE. Elements only

41 45

Quiz

Which one of the following is an extensive property?

A. TemperatureB. MassC. TasteD. DensityE. Color

42 45

Quiz

Which one of the following is a chemical change?

A. Iron �lings are separated from sand using a magnetB. Liquid nitrogen boils to become nitrogen gasC. Gunpowder is explodedD. Antifreeze is added to water in an automobile radiatorE. A shaken cola can is opened producing a spray of soda andcarbon dioxide gas

43 45

Quiz

Which of the following converts chemical energy to mechanicalenergy?

A. A waterwheelB. A kerosene heaterC. A gasoline-powered automobile engineD. A battery attached to a light bulbE. A solar oven

44 45

Problems

When a mixture of aluminum powder and iron(III) oxide isignited, it produces molten iron and aluminum oxide. In anexperiment, 5.40 g of aluminum was mixed with 18.50 g of iron(III)oxide. At the end of the reaction, the mixture contained 11.17 g ofiron, 10.20 g of aluminum oxide, and an undetermined amount ofunreacted iron(III) oxide. No aluminum was left. What is themass of the iron(III) oxide?

Answer: 2.53 g

45 / 45

Problems

When a mixture of aluminum powder and iron(III) oxide isignited, it produces molten iron and aluminum oxide. In anexperiment, 5.40 g of aluminum was mixed with 18.50 g of iron(III)oxide. At the end of the reaction, the mixture contained 11.17 g ofiron, 10.20 g of aluminum oxide, and an undetermined amount ofunreacted iron(III) oxide. No aluminum was left. What is themass of the iron(III) oxide?Answer: 2.53 g

45 / 45

General and Inorganic Chemistry I - Lecture 1nlcd.elte.hu/szalai/pdf/lecture-1-2019.pdfGeneral and Inorganic Chemistry I Lecture 1 IstvÆn Szalai Eötvös University 2019. Outline

Documents