Chemical Reactions

1
Chemical Reactions
2
Balancing Chemical Equations- Problem

• Sodium metal reacts with water to produce aqueous
  sodium hydroxide and hydrogen.

3
Balancing Chemical Equations- Problem

• Sodium metal reacts with water to produce aqueous
  sodium hydroxide and hydrogen.
• Na(s) H2O(l ) ?

4
Balancing Chemical Equations- Problem

• Sodium metal reacts with water to produce aqueous
  sodium hydroxide and hydrogen.
• Na(s) H2O(l ) ? NaOH(aq) H2(g)

5
Balancing Chemical Equations- Problem

• Sodium metal reacts with water to produce aqueous
  sodium hydroxide and hydrogen.
• Na(s) H2O(l ) ? NaOH(aq) H2(g)
• The equation is not yet balanced. Hydrogens come
  in twos on the left, and three hydrogens are on
  the right side of the equation.

6
Balancing Chemical Equations- Problem

• Sodium metal reacts with water to produce aqueous
  sodium hydroxide and hydrogen.
• Na(s) H2O(l ) ? NaOH(aq) H2(g)
• Try a 2 in front of the water.

7
Balancing Chemical Equations- Problem

• Sodium metal reacts with water to produce aqueous
  sodium hydroxide and hydrogen.
• Na(s) 2 H2O(l ) ? NaOH(aq) H2(g)
• We now have two O atoms on the left, so we need
  to put a 2 before NaOH.

8
Balancing Chemical Equations- Problem

• Sodium metal reacts with water to produce aqueous
  sodium hydroxide and hydrogen.
• Na(s) 2 H2O(l ) ?2NaOH(aq) H2(g)
• The two sodium atoms on the right require that we
  put a 2 in front of Na on the left.

9
Balancing Chemical Equations- Problem

• Sodium metal reacts with water to produce aqueous
  sodium hydroxide and hydrogen.
• 2 Na(s) 2 H2O(l ) ?2NaOH(aq) H2(g)
• The two sodium atoms on the right require that we
  put a 2 in front of Na on the left. The equation
  is now balanced.

10
Balancing Chemical Equations- Problem

• Sodium metal reacts with water to produce aqueous
  sodium hydroxide and hydrogen.
• 2 Na(s) 2 H2O(l ) ?2NaOH(aq) H2(g)
• Left Side Right Side
• Na- 2 Na- 2
• H- 4 H- 4
• O- 2 O- 2

11
Chemical Equations

• 2 Na(s) 2 H2O(l ) ?2NaOH(aq) H2(g)
• The balanced chemical equation can be
  interpreted in a variety of ways.
• It could say that 2 atoms of sodium react with
  2 molecules of water to produce 2 molecules of
  sodium hydroxide and a molecule of hydrogen.

12
Chemical Equations

• 2 Na(s) 2 H2O(l ) ?2NaOH(aq) H2(g)
• The balanced chemical equation can be
  interpreted in a variety of ways.
• It could say that 200 atoms of sodium react
  with 200 molecules of water to produce 200
  molecules of sodium hydroxide and 100 molecules
  of hydrogen.

13
Chemical Equations

• 2 Na(s) 2 H2O(l ) ?2NaOH(aq) H2(g)
• The balanced chemical equation can be
  interpreted in a variety of ways.
• It is usually interpreted as 2 moles of sodium
  will react with 2 moles of water to produce 2
  moles of sodium hydroxide and 1 mole of hydrogen.
• The balanced equation tells us nothing about the
  masses of reactants or products.

14
Types of Chemical Reactions

• A combination reaction is a reaction with two or
  more reactants, and a single product.
• 2 CO (g) O2(g) ? 2 CO2(g)

15
Types of Chemical Reactions

• A decomposition reaction is a reaction has one
  reactant, and two or more products.
• CaCO3(s) ? CaO(s) CO2(g)

16
Types of Chemical Reactions

• In a combustion reaction, a substance burns in
  the presence of oxygen. If the reactant is a
  compound containing on carbon, hydrogen and
  oxygen, the products are water and carbon
  dioxide.
• CH4(g) 2 O2(g) ? 2 H2O (l) CO2(g)

17
Chemical Composition

• Usually, the compound is combusted in the
  presence of oxygen. Any carbon in the compound
  is collected as carbon dioxide (CO2), and any
  hydrogen is collected as water (H2O).

18
Determining Empirical Formulas

• If given combustion data
• The ultimate goal is to get the simplest whole
  number ratio of the elements in the compound.
  Usually the compound contains carbon, hydrogen
  and perhaps oxygen or nitrogen.
• 1. Use the information about CO2 to determine
  the moles and mass of carbon in the compound.

19
Determining Empirical Formulas

• If given combustion data
• 2. Use the information about H2O to determine
  the moles and mass of hydrogen in the compound.
• 3. The mass and moles of oxygen (or a third
  element) can be obtained by difference.
• 4. Once moles of each element is obtained, find
  the relative number of moles and empirical
  formula (as with composition).

20
Formulas from Combustion Data
This method assumes the compound contains only C
and H.
21
Empirical Formula using Combustion Data- Problem

• A compound, which contains C, H and O, is
  analyzed by combustion. If 10.68 mg of the
  compound produces 16.01 mg of carbon dioxide and
  4.37 mg of water, determine the empirical formula
  of the compound.
• If the compound has a molar mass of 176.1
  g/mol, determine the molecular formula of the
  compound.

22
Stoichiometry

• Stoichiometry is a Greek word that means using
  chemical reactions to calculate the amount of
  reactants needed and the amount of products
  formed.
• Amounts are typically calculated in grams (or
  kg), but there are other ways to specify the
  quantities of matter involved in a reaction.

23
Stoichiometry

• A balanced chemical equation or reaction is
  needed before any calculations can be made.
• The formulas of all reactants and products are
  written before attempting to balance the equation.

24
Stoichiometry Problem

• Sodium metal reacts with water to produce aqueous
  sodium hydroxide and hydrogen. How many grams of
  sodium are needed to produce 50.0g of hydrogen?
• A balanced chemical equation is needed before
  any calculations can be made.

25
Stoichiometry Problem

• Sodium metal reacts with water to produce aqueous
  sodium hydroxide and hydrogen. How many grams of
  sodium are needed to produce 50.0g of hydrogen?
• Na(s) H2O(l ) ? NaOH(aq) H2(g)

26
Stoichiometry Problem

• Sodium metal reacts with water to produce aqueous
  sodium hydroxide and hydrogen. How many grams of
  sodium are needed to produce 50.0g of hydrogen?
• 2 Na(s) 2 H2O(l ) ? 2 NaOH(aq) H2(g)

27
Stoichiometry Problem

• 2 Na(s) 2 H2O(l ) ?2NaOH(aq) H2(g)
• How many grams of sodium are needed to produce
  50.0g of hydrogen?
• 2 Na(s) 2 H2O(l ) ?2NaOH(aq) H2(g)
• ? grams 50.0 g

28
Stoichiometry Problem

• 2 Na(s) 2 H2O(l ) ?2NaOH(aq) H2(g)
• ? grams 50.0 g
• Although the question doesnt state it, you can
  assume enough water is present for complete
  reaction.
• We can map out the problem
• g H2 ? moles H2? moles Na ? grams Na

29
Stoichiometry Problem

• 2 Na(s) 2 H2O(l ) ?2NaOH(aq) H2(g)
• ? grams 50.0 g
• We can map out the problem
• g H2 ? moles H2? moles Na ? grams Na
• We use the molar mass of H2 to go from grams of
  H2 to moles of H2.

30
Stoichiometry Problem

• 2 Na(s) 2 H2O(l ) ?2NaOH(aq) H2(g)
• ? grams 50.0 g
• We can map out the problem
• g H2 ? moles H2? moles Na ? grams Na
• molar mass of H2

31
Stoichiometry Problem

• 2 Na(s) 2 H2O(l ) ?2NaOH(aq) H2(g)
• ? grams 50.0 g
• We can map out the problem
• g H2 ? moles H2? moles Na ? grams Na
• molar mass of H2
• We use the coefficients from the balanced
  equation to go from moles of H2 to moles of Na.

32
Stoichiometry Problem

• 2 Na(s) 2 H2O(l ) ?2NaOH(aq) H2(g)
• ? grams 50.0 g
• We can map out the problem
• g H2 ? moles H2? moles Na ? grams Na
• molar mass of H2 coefficients

33
Stoichiometry Problem

• 2 Na(s) 2 H2O(l ) ?2NaOH(aq) H2(g)
• ? grams 50.0 g
• We can map out the problem
• g H2 ? moles H2? moles Na ? grams Na
• molar mass of H2 coefficients
• We use the molar mass of Na to go from moles of
  Na to grams of Na.

34
Stoichiometry Problem

• 2 Na(s) 2 H2O(l ) ?2NaOH(aq) H2(g)
• ? grams 50.0 g
• We can map out the problem
• g H2 ? moles H2? moles Na ? grams Na
• molar mass of H2 coefficients molar mass
  of Na

35
Stoichiometry Problem

• 2 Na(s) 2 H2O(l ) ?2NaOH(aq) H2(g)
• ? grams 50.0 g
• g H2 ? moles H2? moles Na ? grams Na
• molar mass of H2 coefficients molar mass
  of Na
• (50.0 g H2) (1 mol H2) (2 moles Na) ( 22.99 g
  Na)
• (2.02 g H2) (1 mol H2) (1 mol Na)

36
Stoichiometry Problem

• 2 Na(s) 2 H2O(l ) ?2NaOH(aq) H2(g)
• (50.0 g H2) (1 mol H2) (2 moles Na) ( 22.99 g
  Na)
• (2.02 g H2) (1 mol H2) (1 mol Na)
• 1,138 grams Na 1.14 x 103 g Na 1.14 kg Na

37
Limiting Reagent Problems

• Sometimes you are given quantities of more than
  one reactant, and asked to calculate the amount
  of product formed. The quantities of reactants
  might be such that both react completely, or one
  might react completely, and the other(s) might be
  in excess. These are called limiting reagent
  problems, since the quantity of one of the reacts
  will limit the amount of product that can be
  formed.

38
Limiting Reagent - Problem

• Aluminum and bromine react to form aluminum
  bromide. If 500. g of bromine are reacted with
  50.0 g of aluminum, what is the theoretical yield
  of aluminum bromide?
• 1. First write the formulas for reactants and
  products.
• Al Br2 ? AlBr3

39
Limiting Reagent - Problem

• Aluminum and bromine react to form aluminum
  bromide. If 500. g of bromine are reacted with
  50.0 g of aluminum, what is the theoretical yield
  of aluminum bromide?
• 2. Now balance the equation by adding
  coefficients.
• 2 Al 3 Br2 ?2 AlBr3

40
Limiting Reagent - Problem

• Aluminum and bromine react to form aluminum
  bromide. If 500. g of bromine are reacted with
  50.0 g of aluminum, what is the theoretical yield
  of aluminum bromide?
• 2 Al 3 Br2 ?2 AlBr3
• The theoretical yield is the maximum amount of
  product that can be formed, given the amount of
  reactants. It is usually expressed in grams.

41
Limiting Reagent - Problem

• 2 Al 3 Br2 ?2 AlBr3
• Given 50.0g 500.g ? grams
• There are several ways to solve this problem.
  One method is to solve the problem twice. Once,
  assuming that all of the aluminum reacts, the
  other assuming that all of the bromine reacts.
  The correct answer is whichever assumption
  provides the smallest amount of product.

42
Limiting Reagent - Problem

• 2 Al 3 Br2 ?2 AlBr3
• Given 50.0g 500.g ? grams
• The problem can be mapped
• Al grams Al ? moles Al? moles AlBr3 ? g AlBr3
• molar mass Al coefficients molar mass
  AlBr3

43
Limiting Reagent - Problem

• 2 Al 3 Br2 ?2 AlBr3
• Given 50.0g 500.g ? grams
• The problem can be mapped
• Al grams Al ? moles Al? moles AlBr3 ? g AlBr3
• molar mass Al coefficients molar mass
  AlBr3
• (50.0 g Al) ( 1 mol Al ) ( 2 mol AlBr3) (266.7
  g AlBr3)
• (26.98 g Al) ( 2 mol Al) (mol
  AlBr3)

44
Limiting Reagent - Problem

• 2 Al 3 Br2 ?2 AlBr3
• Given 50.0g 500.g ? grams
• Al (50.0 g Al) ( 1 mol Al ) ( 2 mol AlBr3)
  (266.7 g AlBr3)
• (26.98 g Al) ( 2 mol Al)
  (mol AlBr3)
• 494. g AlBr3 (if all of the Al reacts)

45
Limiting Reagent - Problem

• 2 Al 3 Br2 ?2 AlBr3
• Given 50.0g 500.g ? grams
• The calculation is repeated for Br2.
• g Br2? moles Br2 ? moles AlBr3 ? g AlBr3
• molar mass Br2 coefficients molar mass
  AlBr3

46
Limiting Reagent - Problem

• 2 Al 3 Br2 ?2 AlBr3
• Given 50.0g 500.g ?
  grams
• The calculation is repeated for Br2.
• g Br2? moles Br2 ? moles AlBr3 ? g AlBr3
• molar mass Br2 coefficients molar mass
  AlBr3
• (500. g Br2) (1 mol Br2) (2 moles AlBr3)
  (266.7 g AlBr3)
• (159.8 g Br2) (3 moles Br2) (1 mol
  AlBr3)
• 556. grams of AlBr3

47
Limiting Reagent - Problem

• 2 Al 3 Br2 ?2 AlBr3
• Given 50.0g 500.g ? grams
• Summary
• We have enough Al to produce 494. g AlBr3
• We have enough Br2 to produce 556. grams of
  AlBr3
• The theoretical yield is 494. grams of AlBr3.

48
Limiting Reagent - Problem

• 2 Al 3 Br2 ?2 AlBr3
• Given 50.0g 500.g 494. g
• Summary
• All of the Al reacts, so Al is limiting.
• Bromine is in excess.
• Additional questions
• 1. How much bromine is left over?
• 2. If 418 grams of AlBr3 is obtained, what is
  the yield?

49
Limiting Reagent - Problem

• 2 Al 3 Br2 ?2 AlBr3
• Given 50.0g 500.g 494. g
• 1. How much bromine is left over?
• Since all 50.0 g of the Al reacts, the product
  must contain 494.g -50.g 444. g of bromine.
  Therefore, 500.g- 444.g 56. g of Br2 are left
  over.

50
Limiting Reagent - Problem

• 2 Al 3 Br2 ?2 AlBr3
• Given 50.0g 500.g 494. g
• If 418 grams of AlBr3 is obtained, what is the
  yield?
• The percent yield is (actual yield) (100)
• (theoretical yield)
• yield (418 g) (100) 84.6
• (494g)

51
Trends in Reactivity of Main Group Elements

• In using an elements position on the periodic
  table to predict reactivity, there are two
  underlying factors to consider.
• The smallest member of a family or group often
  exhibits different behavior than the rest of the
  family due to its small size.
• Elements on a diagonal often exhibit similar
  behavior even though they are in different
  groups.

52
Trends in Reactivity of Main Group Elements

• For example, both Li and Mg react with nitrogen
  to form nitrides (Li3N and Mg3N2), although
  nitrogen is inert with most elements.

53
Metallic Character
54
Hydrogen

• Hydrogen is a non metal. It bonds covalently,
  with the only ionic form being the hydride ion,
  H. In this way it is similar to the halogens.
  And is sometimes placed in both group IA and
  group VIIA on periodic tables.

55
Metallic Character

• Across a period, metallic behavior decreases.
  Non-metals are often crumbly solids, liquids or
  gases at room temperature.

56
Metallic Character

• Metallic behavior increases going down a group.