Notes

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For the following balanced equation: 3 Cu(s) + 8 HNO3(aq) → 3 Cu(NO3)2(aq) + 2 NO(g) + 4 H2O(l) a) How many moles of HNO3 will react with 10 moles of Cu? b) How many moles of NO will form if 0.50 moles of Cu reacts? c) If 0.80 moles of H2O forms, how much NO must also form? d) How many moles of Cu are in 10.0 grams of Cu? e) If 10.0 g of Cu reacts, how many moles of NO will form? f) If 10.0 g of Cu reacts, how many grams of HNO3 are required? g) If 10.0 g of Cu and 20.0 g of HNO3 are put together in a reaction vessel, which one will be in excess? h) How many grams of the excess substance will be left over? i) How many grams of NO will form in the reaction described in part g?

a) 26.67 moles HNO3

b) 0.33 moles NO

c) 0.40 moles NO is produced

d).157 moles Cu

e) 0.105 moles NO

f) 26.4 grams HNO3

g) Cu is in excess

h) 2.41 grams Cu remain

i) 2.37 grams NO

Explanation:

Step 1: Data given

Molar mass of Cu = 63.55 g/mol

Molar mass of HNO3 = 63.01 g/mol

Molar mass of Cu(NO3)2 = 187.56 g/mol

Molar mass of NO = 30.01 g/mol

Molar mass of H2O = 18.02 g/mol

Step 2: The balanced equation

3 Cu(s) + 8 HNO3(aq) → 3 Cu(NO3)2(aq) + 2 NO(g) + 4 H2O(l)

a) How many moles of HNO3 will react with 10 moles of Cu?

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 10 moles Cu we need 8/3 *10 = 26.67 moles HNO3

b) How many moles of NO will form if 0.50 moles of Cu reacts?

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 0.50 moles Cu we'll have 2/3 *0.50 = 0.33 moles NO

c) If 0.80 moles of H2O forms, how much NO must also form?

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

If 0.80 moles H2O is produced, 0.80/2 = 0.40 moles NO is produced

d) How many moles of Cu are in 10.0 grams of Cu?

Moles Cu = 10.0 grams / 63.55 g/mol = 0.157 moles

In 10.0 grams Cu we have 0.157 moles Cu

e) If 10.0 g of Cu reacts, how many moles of NO will form?

10.0 grams Cu = 0.157 moles

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 0.157 moles Cu we'll have 2/3 * 0.157 = 0.105 moles NO

f) If 10.0 g of Cu reacts, how many grams of HNO3 are required?

10.0 grams Cu = 0.157 moles

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 0.157 moles Cu we'll need 0.419 moles HNO3

This is 0.419 moles * 63.01 g/mol = 26.4 grams HNO3

g) If 10.0 g of Cu and 20.0 g of HNO3 are put together in a reaction vessel, which one will be in excess?

Moles Cu = 0.157 moles

Moles HNO3 = 20.0 grams / 63.01 g/mo = 0.317 moles

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

The limiting reactant is HNO3. It will completely be consumed (0.317 moles). Cu is in excess. There will react 3/8 * 0.317 = 0.119 moles Cu

There will remain 0.157 - 0.119 = 0.038 moles

h) How many grams of the excess substance will be left over?

There will react 3/8 * 0.317 = 0.119 moles Cu

There will remain 0.157 - 0.119 = 0.038 moles

This is 0.038 moles * 63.55 g/mol = 2.41 grams

i) How many grams of NO will form in the reaction described in part g?

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 0.317 moles HNO3 we'll have 0.317/4 = 0.0793 moles NO

This is 0.079 mol * 30.01 g/mol = 2.37 grams NO

BALANCED ALREADY?

2

Indicate what type, or types, of reaction each of the following represents:

Ca(s)+Br2(l)→CaBr2(s)Ca(s)+Br2(l)→CaBr2(s)

Ca(OH)2(aq)+2HBr(aq)→CaBr2(aq)+2H2O(l)Ca(OH)2(aq)+2HBr(aq)→CaBr2(aq)+2H2O(l)

C6H12(l)+9O2(g)→6CO2(g)+6H2O(g)C6H12(l)+9O2(g)→6CO2(g)+6H2O(g)

Answer a

oxidation-reduction (addition)

Answer b

acid-base (neutralization)

Answer c

oxidation-reduction (combustion)

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Balancing chemical equations notes

1. Balancing Chemical Equations TEKS 8.5D recognize that chemical formulas are used to identify substances and determine the number of atoms of each element in chemical formulas containing subscripts TEKS 8.5F recognize whether a chemical equation containing coefficients is balanced or not and how that relates to the law of conservation of mass

2. occurs when the chemical bonds between elements are formed or broken. A chemical reaction

3. This requires a gain or loss of energy.

4. is a short, simple way to describe a chemical reactions using symbols and formulas instead of words. A chemical equation

5. are the substances you begin with in a chemical reaction. Reactants

6. These are shown to the left side of an arrow called the yield sign. Fe + 2 HCl FeCl2 + H2 reactants products yield sign

7. During a chemical reaction, the atoms rearrange to form new bonds between atoms.

8. Products are the substances you end with in a chemical reaction.

9. Products are shown to the right of the yield sign arrow. Fe + 2 HCl FeCl2 + H2 reactants products yield sign

10. The products formed in a chemical reaction have different properties from the reactants.

11. A yield sign Fe + 2 HCl FeCl 2 + H 2 is used to show the direction of the chemical reaction.

12. A chemical equation is read like this: 1 atom of iron reacts with 2 molecules of hydrochloric acid to yield 1 molecule of Iron chloride and 1 molecule of hydrogen Fe + 2 HCl FeCl2 + H2

13. During a chemical reaction, atoms cannot be created or destroyed, so the number of atoms at the beginning of a reaction MUST equal the number at the end of the reaction!

14. This is called the Law of Conservation of Mass.

15. This law also means that the mass of the reactants must equal the mass of the products.

16. What if the number of atoms in the reactants does not match the number in the products?

17. C3H8 + O2 CO2 + H2O mass =76 gramsmass =76 grams mass = 62 gramsmass = 62 grams 3-carbon3-carbon 1-carbon1-carbon 8-hydrogen8-hydrogen 2-hydrogen2-hydrogen 2-oxygen2-oxygen 3-oxygen3-oxygen

18. The Law of Conservation of Mass says that the number of each atom must be equal on both sides of the yield sign along with the masses. Let’s balance the equation using some basic rules.

19. Basic Rules for Balanced Equations

20. Chemical formulas CANNOT be altered!

21. When balancing an equation, the subscript cannot be changed. H2O Don’t change me!

22. Coefficients can be added before an element or molecule.

23. C3H8 + O2 => CO2 + H2O5 3 4 Add coefficients to the equation to balance the number of atoms

24. C3H8 + 5O2 => 3CO2 + 4H2O 3-carbon 3-carbon 8-hydrogen 8- hydrogen 10-oxygen 10- oxygen Atoms on each side are equal!!

25. Coefficients were used to balance the number of atoms on each side of the chemical equation.

26. If we mass the reactants and products before and after the reaction, the masses are equal. We have followed the law! mass = 204 grams =>mass = 204 grams C3H8 + 5O2 => 3CO2 + 4H2O

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ATOMS

3. Atomic Structure All matter is composed of atoms. Understanding the structure of atoms is critical to understanding the nature of biological molecules. 5

4. Atomic Structure Atoms are composed of -protons – positively charged particles -neutrons – neutral particles -electrons – negatively charged particles Protons and neutrons are located in the nucleus. Electrons are found in orbitals surrounding the nucleus. 6

5. Atomic Structure Every different atom has a characteristic number of protons in the nucleus. atomic number = number of protons Atoms with the same atomic number have the same chemical properties and belong to the same element. 7

6. Atomic Structure Each proton and neutron has a mass of approximately 1 dalton. The sum of protons and neutrons is the atom’s atomic mass. Isotopes – atoms of the same element that have different atomic mass numbers due to different numbers of neutrons. 8

7. Electrons Protons Neutrons Hydrogen Helium

8. Carbon Oxygen Phosphorous Calcium

9. Ions Loss or gain of an electron makes an atom into an ion.

10. Isotopes Atoms that gain or lose a neutron become isotopes. Radioactive isotopes are used in medicine for imaging (such as PET scanners).

11. Ionic Bonding

12. Ionic substances tend to form crystaline lattices rather than distinct molecules.

13. Covalent Bonding Co = together valent = valence shells The hydrogen atom.

14. Covalent Bonding

15. Polar bonding between atoms produces a polar molecule, which has areas with slightly positive or slightly negative charges.

16. polarity Salt dissolves in water. Oil does not.

17. Molecules bond together Weak attractive forces between molecules are called hydrogen bonds.

18. Here, sugar (a polar substance) is in solution with water (also polar). Mark where the hydrogen bonds will form.

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2. TOPIC ORGANOPLATINUM CHEMISTRY

3. CONTENT  INTRODUCTION  ORGANOPLATINUM COMPOUNDS i. ORGANOPLATINUM (0) ii. ORGANOPLATINUM (I) iii. ORGANOPLATINUM (II) iv. ORGANOPLATINUM (IV)  REFERENCES

4. ORGANOPLATINUM ORGANOPLATINUM CHEMISTRY IS A CHEMISTRY OF ORGANOMETALLIC COMPOUNDS CONTAINING A CARBON TO PLATINUM BOND AND THE STUDY OF PLATINUM AS A CATALYST

5. - ORGANOPLATINUM COMPONDS EXIST IN THE OXIDATION STATE 0 TO IV ,WITH OXIDATION STATE (II) MOST ABUNDANT

6. ORGANOPLATINUM (0)  MOST ORGANOPLATINUM (0) COMPOUNDS CONTAIN ALKENE AND ALKYNE LIGANDS.  THE ALKENE AND ALKYNE LIGANDS SERVE AS TWO ELECTRON DONORS.  Eg: (PPh₃)₂Pt(C₂H₄) AND (PPh₃)₂Pt(C₂Ph₂) .  NITROGEN BASED LIGANDS DO NOT OFTEN SUPPORT THE FORMATION OF PLATINUM COMPLEXES OF ALKENES AND ALKYNES .

7. ORGANOPLATINUM (1)  PLATINUM (I) COMPOUNDS ARE UNCOMMON BUT GENERALLY ARE DIMAGNETIC BECAUSE THEY HAVE Pt-Pt BONDS.  Eg : [Pt₂(CO)₆]

8. ORGANOPLATINUM (II)  A HISTORICALLY SIGNIFICANT ORGANOPLATINUM(II) COMPOUND IS ZEISE’S SALT ,WHICH IS OBTAINED FROM ETHYLENE AND POTASSIUM TETRACHLOROPLATINATE.  PLATINUM ALLYL COMPLEXES ARE ALSO COMMON .

9. ORGANOPLATINUM (IV)  THE FIRST ORGANOPLATINUM COMPOUND EVER SYNTHESISED WAS TRIMETHYL PLATINUM IODIDE FROM PLATINUM (IV) CHLORIDE AND METHYL MAGNESIUM IODIDE  THE COMPOUND ADOPTS A CUBANE LIKE STRUCTURE WITH FOUR TRIPLY BRIDGING IODIDE LIGANDS .

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