Title: Chemical Reactions
1Chemical Reactions
2Balancing Chemical Equations- Problem
- Sodium metal reacts with water to produce aqueous
sodium hydroxide and hydrogen. -
3Balancing Chemical Equations- Problem
- Sodium metal reacts with water to produce aqueous
sodium hydroxide and hydrogen. - Na(s) H2O(l ) ?
4Balancing Chemical Equations- Problem
- Sodium metal reacts with water to produce aqueous
sodium hydroxide and hydrogen. - Na(s) H2O(l ) ? NaOH(aq) H2(g)
5Balancing 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.
6Balancing 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.
7Balancing 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.
8Balancing 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.
9Balancing 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.
10Balancing 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
11Chemical 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.
12Chemical 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.
13Chemical 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.
14Types 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)
15Types of Chemical Reactions
- A decomposition reaction is a reaction has one
reactant, and two or more products. - CaCO3(s) ? CaO(s) CO2(g)
-
16Types 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)
-
17Chemical 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).
18Determining 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. -
19Determining 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).
20Formulas from Combustion Data
This method assumes the compound contains only C
and H.
21Empirical 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.
22Stoichiometry
- 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.
23Stoichiometry
- 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.
24Stoichiometry 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.
25Stoichiometry 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)
26Stoichiometry 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)
27Stoichiometry 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
28Stoichiometry 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
29Stoichiometry 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.
30Stoichiometry 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
31Stoichiometry 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.
32Stoichiometry 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
33Stoichiometry 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.
34Stoichiometry 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
35Stoichiometry 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)
36Stoichiometry 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
37Limiting 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.
38Limiting 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
39Limiting 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
40Limiting 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.
41Limiting 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. -
42Limiting 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
43Limiting 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)
44Limiting 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)
45Limiting 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
46Limiting 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
47Limiting 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.
-
48Limiting 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?
49Limiting 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. -
50Limiting 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)
51Trends 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.
52Trends 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.
53Metallic Character
54Hydrogen
- 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.
55Metallic Character
- Across a period, metallic behavior decreases.
Non-metals are often crumbly solids, liquids or
gases at room temperature.
56Metallic Character
- Metallic behavior increases going down a group.