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Like a recipe: Reactants Products 2H 2 (g) + O 2 (g)  2H 2 O(l) coefficients subscripts - PowerPoint PPT Presentation


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Balancing Eqns. Like a recipe: Reactants Products 2H 2 (g) + O 2 (g)  2H 2 O(l) coefficients subscripts. Balancing Eqns. Symbols  Yields or Produces (s) solid (l) liquid (pure liquid) (aq) aqueous (dissolved in water) (g) or  gas. Balancing.

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Balancing Eqns

Like a recipe:

ReactantsProducts

2H2(g) + O2(g)  2H2O(l)

coefficients subscripts


Balancing Eqns

Symbols

 Yields or Produces

(s) solid

(l) liquid (pure liquid)

(aq) aqueous (dissolved in water)

(g) or  gas


Balancing

Copper(II) Chloride reacts with Iron(III) Sulfate to form Copper(II)Sulfate and Iron(III) Chloride

Aluminum nitrate reacts with Sodium hydroxide to form Aluminum hydroxide and Sodium nitrate


Some types of reactions
Some Types of Reactions

1. Synthesis

Al + Cl2 AlCl3

CaO(s) + CO2(g)  CaCO3(s)

  • Decomposition

    HgO(s)  Hg(l) + O2(g)

    CaCO3(s)  CaO(s) + CO2(g)


Some types of reactions1
Some Types of Reactions

  • Combustion

    Fe + O2 Fe2O3

    C4H10 + O2  CO2 + H2O

  • Other Types

    Single Replacement

    Double Replacement



Stoichiometry

Mole

1 dozen = 12 items

1 mole = 6.022 X 1023 atoms/molecules

1 gram hydrogen = 6.022 X 1023 atoms of hydrogen


Stoichiometry

Grams Moles Atoms

1 g H 1 mole H 6.02 X 1023 atms

2 g H

12 g C

36 g C

48 g O


Stoichiometry

Grams Moles Molecules Atoms

16 g CH4

8 g CH4

88 g CO2

131 g Ba(NO3)2


Molar Mass

1. Molar Mass = mass of one mole

2. Element= atomic mass

1 mole of O = 16.0 grams

3. Molecule or Compound – sum of all the atoms


Molar Mass

Calculate the molar mass of

Barium

O2

BaCl2?

Fe2(SO4)3?


GMA

Grams

Moles

Atoms

  • How many C atoms are present in 18.0 g? (Ans: 9.03 X 1023 C)

  • What is the mass of 1.20 X 1024 atoms of Na? (Ans: 45.8 grams)


GMA

3. What is the mass of 1.51 X 1023 atoms of Be?(Ans: 2.26 g)

5. How many atoms and grams are in 0.400 mol of Radium? (Ans: 90.4 g, 2.41 X 1023 atoms)


GMMA

1. Monoatomic Elements (C, Fe, Au)

GMA

2. Molecules and Ionics (H2O, CaCl2, O2)

GMMA


GMMA

3. Molecules and formula units work the same when converting

Molecules = Molecular Comps

Formula Units = Ionic Compounds

G

M

M

A


GMMA

  • How many calcium and chlorine atoms are in 200.0 grams of Calcium Chloride?

    (Ans: 2.17 X 1024 atoms Cl)


GMMA

2. How many hydrogen and oxygen atoms are in 3.60 grams of H2O? (Ans: 2.41 X 1023 atoms H)


GMMA

  • Given 3.01X1024 molecules of SO3, find everything else.

  • Given 3.01 X 1022 molecules of Iron(III)bromide, find everything else.


Mixed Examples

1. How many carbon atoms are in 36.0 grams of carbon (1.81 X 1024)

2. How many carbon atoms are in 36.0 grams of C2H6? (Ans:1.45 X 1024 atoms of C)


Mixed Examples

Homework Problems (find everything else)

a) 10.0 g C

b) 10.0 g C2H6

c) 4.0 X 1023 atoms of S

d) 4.0 X 1023 molecules of SO2

e) 0.44 moles of SO2


Empirical Formula

1. Empirical formula - simplest ratio of the elements in a compound

2. Formula Empirical Form.

C2H2

Al4S6

C6H12O6

C12H24O12


Empirical Formula

1. What is the EF of a compound that has 0.900 g Ca and 1.60 g Cl?

Rules

- Go to moles

- Divide by the smaller


Empirical Formula

2. What is the EF of a compound that is 66.0 % Ca and 34.0% P?

3. What is the EF of a compound that is 43.7 % P and 56.3 % O?


Molecular Formula

1. Empirical –ratios of the elements

2. Molecular –true number of each element


Molecular Formula

EFMF

CH2O CH2O (30 g/mol)

C2H4O2 (60 g/mol)

C3H6O3 (90 g/mol)

C4H8O4 (120 g/mol)


Molecular Formula

1. What is the MF of benzene if it has an EF of CH and a molar mass of 78.0 g?

2. What is the MF of a compound that is 40.9% C, 4.58 % H and 54.5 % O? It has a molar mass between 350 and 360 g/mol.


Reaction Stoich.

What coefficients mean:

2 Na + Cl2 2NaCl

2 Na 1 Cl2 2NaCl

4 Na

6 Na


Reaction Stoich.

2 Na + Cl2 2NaCl

4 Cl2

2 moles Na

10 moles Na

ONLY WORKS FOR MOLES AND MOLECULES


Reaction Stoich.

1. How many moles of H2 and O2 must react to form 6 moles of H2O?

2. How many moles of KCl and O2 are formed from the decomposition of 6 moles of KClO3?


Reaction Stoich.

  • How many grams of oxygen are needed to react with 14.6 g of Na to form Na2O? (Ans: 5.08 g)

  • How many grams of P4 and O2 are needed to make 3.62 g of P2O5? (Ans: 1.58 g, 2.04 g)


Reaction Stoich.

5. What mass of oxygen is needed to react with 16.7 g of iron to form Iron(III)oxide? (Ans: 7.18 g)

4Fe + 3O2 2Fe2O3


Calculate the mass of sodium bromide and oxygen that are formed from the decomposition of 50.0 grams of sodium bromate (NaBrO3).

(34.1 g NaBr, 15.9 g O2)


Limiting Reactant formed from the decomposition of 50.0 grams of sodium bromate (NaBrO

  • Sandwich analogy:

    13 slices of bread

    4 pieces of turkey

    Maximum # of sandwiches?

    2. Limiting Reactant – Totally consumed in a reaction. No leftovers


Limiting Reactant formed from the decomposition of 50.0 grams of sodium bromate (NaBrO

  • How many grams of H2SO4 can be formed from the rxn of 5.00 moles of SO3 and 2.00 moles of H2O?

    SO3 + H2O  H2SO4

    (Ans: 196 g)


Limiting Reactant formed from the decomposition of 50.0 grams of sodium bromate (NaBrO

  • How many grams of H2O can be formed from the rxn of 6.00 moles of H2 and 4.00 moles of O2?

    O2 + H2 H2O


Limiting Reactant formed from the decomposition of 50.0 grams of sodium bromate (NaBrO

2. How many grams of NaCl can be formed from the reaction of 0.300 mol of Na and 0.100 mol of Cl2?

2Na + Cl2 2NaCl

(Ans: 11.7 g)


Limiting Reactant formed from the decomposition of 50.0 grams of sodium bromate (NaBrO

  • How many grams of Ag can be formed from the rxn of 2.00 g of Zn and 2.50 g of silver nitrate? How much excess reactant remains?

    Zn + AgNO3 Ag + Zn(NO3)2

    (Ans: 1.59 g Ag, 1.52 g xs zinc)


Limiting Reactant formed from the decomposition of 50.0 grams of sodium bromate (NaBrO

  • How many grams of Ba3(PO4)2 can be formed from the rxn of 3.50 g of Na3PO4 and 6.40 g of Ba(NO3)2?

    Na3PO4 + Ba(NO3)2 Ba3(PO4)2 + NaNO3

    (Ans: 4.92 g)


Limiting Reactant formed from the decomposition of 50.0 grams of sodium bromate (NaBrO

  • How many grams of Ag2S can be formed from the rxn of 15.6 g of Ag and 2.97 g of H2S? (Assume O2 is in excess)

    4Ag + 2H2S + O2 2Ag2S + 2H2O

    (Ans: 18.1 g)


Percent Yield formed from the decomposition of 50.0 grams of sodium bromate (NaBrO

  • Formula:

    Actual Yield X 100 = % Yield

    Theoretical Yield


Percent Yield formed from the decomposition of 50.0 grams of sodium bromate (NaBrO

  • What is the % yield if you start with 10.00 grams of C and obtain 1.49 g of H2 gas?

    C + H2O  CO + H2

    (Ans: 89.4%)


Percent Yield formed from the decomposition of 50.0 grams of sodium bromate (NaBrO

  • Carbon was heated strongly in sulfur(S8) to form carbon disulfide. What is the percent yield if you start with 13.51 g of sulfur and collect 12.5 g of CS2?

    4C + S8 4CS2

    (Ans: 78.0%)


Percent Yield formed from the decomposition of 50.0 grams of sodium bromate (NaBrO

3. 36.7 grams of CO2 were formed from the rxn of 40.0 g of CH3OH and 46.0 g of O2. What is the % yield?

2CH3OH + 3O2 2CO2 + 4H2O

(ANS: 87%)


In this experiment, magnesium chloride was prepared and its empirical formula was compared to the accepted formula of MgCl2. To prepare magnesium chloride, 0.40 grams of magnesium powder was combined with 10 mL of 0.10 M HCl. The mixture was allowed to react, and heated to dryness. The mass of the resulting crystals was used to calculate the empirical formula. The average calculated formula of MgCl1.8 had a 10% error and a range was 0.40 chlorine atoms. This procedure was not effective because while it was accurate, it was not precise.


8a) SO empirical formula was compared to the accepted formula of MgCl3 + H2O  H2SO4

  • B2S3 + 6H2O  2H3BO3 + 3H2S

  • 4PH3 + 8O2 6H2O + P4O10

  • 2Hg(NO3)2 2HgO + 4NO2 + O2

  • Cu + 2H2SO4  CuSO4 + SO2 + 2H2O


  • a) 6 1 2 14. a) 1 1 1 empirical formula was compared to the accepted formula of MgCl

    b) 1 3 2 b) 1 6 2 3

    c) 2 2 1 4 c) 1 2 2 1

    d) 1 6 3 2 d) 2 2 4 1

    e) 3 2 1 6 e) 1 2 1 1 2

    f) 2 1 1 2

    g) 4 9 4 10 2


  • a) 4Al + 3O empirical formula was compared to the accepted formula of MgCl2 2Al2O3

    b) Cu(OH)2  CuO + H2O

    c) C7H16 + 11O2  7CO2 + 8H2O

    d) 2C5H12O + 15O2  10CO2 + 12H2O

  • 2 9 6 6

    1 1 2

    1 6 5 3

    1 3 2

    1 1 2

  • a) 44.0 g/mol b) 122.0 g/mol

    c) 58.3 g/mol d) 60.0 g/mol e) 130.0 amu


24. 26.0 g/mole 92.3% C empirical formula was compared to the accepted formula of MgCl

176.0 g/mole 4.5% H

132.1 g/mole 6.1% H

300.1 g/mole 65.01% Pt

272.0 g/mole 11.8% O

305.0 g/mole 70.8 % C


  • a)K empirical formula was compared to the accepted formula of MgCl3PO4 b) Na2SiF6 c) C12H12N2O3

  • a) H2C2O4 b) C4H8O2

  • a) C13H18O2 b) C5H14N2 c) C9H13O3N


58.a) 0.800 mol CO empirical formula was compared to the accepted formula of MgCl2 b) 14.7 g C6H12O6

c) 7.16 g CO2

60. a) 0.939 mol Fe2O3 b) 78.9 g CO

c) 105 g Fe d) 229 g= 229 g


  • a) H empirical formula was compared to the accepted formula of MgCl2C2O4 b) C4H8O2

    50. a) C13H8O2 b)C5H14N2

    c) C9H13O3N

  • a) 0.800 mol CO2 b) 14.7 g C6H12O6

    c) 7.18 g CO2

  • a) Fe2O3 + 3CO  2Fe + 3CO2

    b) 78.9 g CO2

    c) 124 g CO2

    d) 229 g = 229 g


  • a) CaH empirical formula was compared to the accepted formula of MgCl2 + 2H2O  Ca(OH)2 + 2H2

    b) 88.75 g CaH2

    64. a) 15.6 mol O2 b) 35.0 g O2 c) 9175.1 g

  • 0.167 mol Al2(SO4)3 form

    0.333 mol Al(OH)3 react

    0.167 mol AL(OH)3 remain

  • a) O2 is limiting reactant

    b) 1.86 g H2O produced

    c) 0.329 g NH3 remain

    d) 4.25 g = 4.25 g


  • 5.24 g H empirical formula was compared to the accepted formula of MgCl2SO4 6.99 g PbSO4

    2.77 g HC2H3O2

  • C2H6 + Cl2 C2H5Cl + HCl

    232 g C2H5Cl (theoretical yield)

    88.8% yield

    80. Actual yield of Na2S = 1.80 g (1.95 g is the theoretical yield)


The atmosphere of Jupiter is composed almost entirely of hydrogen (H2) and helium (He). If the average molar mass of Jupiter’s atmosphere is 2.254 g/mole, calculate the percent composition.


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