Transcript Stoichiometry Notes

GOOD AFTERNOON!
Prepare to take notes. You will be allowed to use
these notes on the test next week! You will need:
1. Something to write on
2. Something to write with
3. A periodic table
4. A calculator (you may have to share with your
neighbor)
STOICHIOMETRY
Calculations using chemical reactions
MOLE-TO-MOLE RATIO
• The coefficients of a balanced chemical reaction can be
used to show the relationship between the products and or
reactants of a chemical reaction. This is true because the
units of a balanced chemical reaction are moles.
• The mole-to-mole ratio is a ratio between reactants and
products which can be used to convert between any two
compounds in a chemical reaction.
• The mole-to-mole ratio is used in any problem that converts
between amounts of two different substances (compounds
or elements).
MOLE-TO-MOLE CONVERSION
PROBLEM EXAMPLE:
According to the following balanced chemical equation, how
many moles of Na2SO4 can be produced from the reaction of 14.3
moles of NaOH with excess H2SO4?
H2SO4 + 2 NaOH  Na2SO4 + 2 H2O
14.3 moles NaOH ? moles Na2SO4
14.3 moles NaOH
1 mole Na2SO4
2 moles NaOH
= 14.3 moles Na2SO4 = 7.15 moles Na2SO4
2
GRAM-TO GRAM STOICHIOMETRY:
• Because the units of chemical reaction coefficients are
moles and not grams, you must always convert to moles
before converting from units of one compound to units of
another compound
GRAM-TO-GRAM STOICHIOMETRY
EXAMPLE PROBLEM:
Example:
How many grams of NaOH are required to react completely with 28.2
grams of H2SO4 according to the following balanced reaction?
H2SO4 + 2 NaOH  Na2SO4 + 2 H2O
28.2 g H2SO4
1 mol H2SO4
2 mol NaOH
40.0 g NaOH
98.1 g H2SO4
1 mol H2SO4
1 mol NaOH
= 23.0 g NaOH
GRAM-TO-MOLE STOICHIOMETRY
EXAMPLE PROBLEM:
Example:
How many moles of NaOH are required to react completely with 28.2
grams of H2SO4 according to the following balanced reaction?
H2SO4 + 2 NaOH  Na2SO4 + 2 H2O
28.2 g H2SO4
1 mol H2SO4
2 mol NaOH
98.1 g H2SO4
1 mol H2SO4
= 0.575 moles NaOH
MOLE-TO-GRAM STOICHIOMETRY
EXAMPLE PROBLEM:
Example:
How many grams of NaOH are required to react completely with 14.3
moles of H2SO4 according to the following balanced reaction?
H2SO4 + 2 NaOH  Na2SO4 + 2 H2O
14.3 mol H2SO4
2 mol NaOH
40.0 g NaOH
1 mol H2SO4
1 mol NaOH
= 1,144 g NaOH
LIMITING REACTANT, EXCESS
REACTANT, AND THEORETICAL YIELD:
• The limiting reactant is the reactant that runs out first. It limits
the reaction more than any other reactant.
• Excess reactants are the reactants that are not used up in a
chemical reaction. They are left over. By definition, they
are all reactants other than the limiting reactant.
• Theoretical yield is the amount of product you should be
able to make using the given amount of each reactant. It is
limited by the limiting reactant and is determined through
calculations using the limiting reactant.
MOLE-TO-MOLE LIMITING
REACTANT PROBLEM:
According to the following balanced chemical equation, how
many moles of Na2SO4 can be produced from the reaction of 14.3
moles of NaOH with 12.4 moles of H2SO4? What is the limiting
reactant, excess reactant, and theoretical yield?
H2SO4 + 2 NaOH  Na2SO4 + 2 H2O
14.3 moles NaOH
1 mole Na2SO4
2 moles NaOH
12.4 moles H2SO4
1 mole Na2SO4
1 mole H2SO4
= 7.15 moles Na2SO4
= 12.4 moles Na2SO4
GRAM-TO-GRAM LIMITING
REACTANT PROBLEM:
According to the following balanced chemical equation, how
many grams of Na2SO4 can be produced from the reaction of 24.3
grams of NaOH with 22.4 grams of H2SO4? What is the limiting
reactant, excess reactant, and theoretical yield?
H2SO4 + 2 NaOH  Na2SO4 + 2 H2O
24.3 g NaOH
22.4 g H2SO4
1 mol NaOH
1 mol Na2SO4
142.1 g Na2SO4
40.0 g NaOH
2 mol NaOH
1 mol Na2SO4
1 mol H2SO4
1 mol Na2SO4
142.1 g Na2SO4
98.1 g H2SO4
1 mol H2SO4
1 mol Na2SO4
= 43.2 g Na2SO4
= 32.4 g Na2SO4
HOW MUCH EXCESS IS LEFT OVER?
1. Determine which reagent is the limiting reagent
2. Find out how much of the excess reagent was used in the
reaction
3. Subtract the amount used (from step 2) from the amount
available (given in the problem).
GRAM-TO-GRAM EXCESS
REACTANT PROBLEM:
According to the following balanced chemical equation, how many
grams of Na2SO4 can be produced from the reaction of 24.3 grams of
NaOH with 22.4 grams of H2SO4? What is the limiting reactant, excess
reactant, and theoretical yield? How much excess reagent is left over?
Step 1:
24.3 g NaOH
22.4 g H2SO4
H2SO4 + 2 NaOH  Na2SO4 + 2 H2O
1 mol NaOH
1 mol Na2SO4
142.1 g Na2SO4
40.0 g NaOH
2 mol NaOH
1 mol Na2SO4
1 mol H2SO4
1 mol Na2SO4
142.1 g Na2SO4
98.1 g H2SO4
1 mol H2SO4
1 mol Na2SO4
= 43.2 g Na2SO4
= 32.4 g Na2SO4
GRAM-TO-GRAM EXCESS
REACTANT PROBLEM:
According to the following balanced chemical equation, how many
grams of Na2SO4 can be produced from the reaction of 24.3 grams of
NaOH with 22.4 grams of H2SO4? What is the limiting reactant, excess
reactant, and theoretical yield? How much excess reagent is left over?
H2SO4 + 2 NaOH  Na2SO4 + 2 H2O
Step 2:
22.4 g H2SO4
1 mol H2SO4
2 mol NaOH
40.0 g NaOH
98.1 g H2SO4
1 mol H2SO4
1 mol NaOH
= 18.3 g NaOH used
Step 3:
24.3 g NaOH (available) -18.3 g NaOH (used) = 6.00 g NaOH left
PERCENT YIELD
• Percent yield is a way to know how efficient a chemical
reaction is by comparing the actual results of a reaction to
the predicted yield (also known as theoretical yield) of a
reaction.
Actual Yield
• The formula for percent yield is % yield =
x100
Theoretical Yield
• Actual yield will be given in the problem on a test or quiz
PERCENT YIELD EXAMPLE
PROBLEM
For the balanced equation shown below, if the reaction of 20.7
grams of CaCO3 produces 6.81 grams of CaO, what is the
percent yield?
CaCO3  CaO+CO2
20.7 g CaCO3
% yield =
1 mol CaCO3
1 mol CaO
56.1 g CaO
100.1 g CaCO3
1 mol CaCO3
1 mol CaO
6.81 g CaO (actual)
11.6 g CaO (theoretical)
x 100 = 58.7%
= 11.6 g CaO