Transcript 4.06 Stoichiometry Stoichiometry • It’s a process that allows us to mathematically convert and calculate the relationships between the amount of reactants and products in.

4.06
Stoichiometry
Stoichiometry
• It’s a process that allows us to
mathematically convert and calculate
the relationships between the amount of
reactants and products in a chemical
reaction.
Graphical representation
Example:
for every O2 molecule,
there are 2 H2O
molecules
2 H2 + 1 O2  2 H2O
Solving stoichiometry equations
It is just like solving an usual math equation. Just to let
you know, stoichiometry equations start with one
substance and ends with another substance.
1. Balance the unequal equation by adding
coefficients.
2. Identify what you have been asked to find in the
question.
3. Set up your equation with the given information you
have, which should be the measure and unit of a
substance.
4. Solve by multiplying everything in the numerator and
dividing by everything in the denominator.
5. Oh yeah, and don’t forget to check your work!
Show me Stoichiometry
It is important that you always keep a
periodic table handy while solving
stoichiometry problems. You will also
encounter stoichiometry questions that
ask you for limiting reactants and
percent yield, but don’t be intimidated.
These are quite easy to calculate as
well.
Now lets solve a few.
Problem #1
How many moles of aluminum hydroxide are needed to react
completely with 12.3 moles of nitric acid?
Al(OH)3 + 3 HNO3 → 3 H2O + Al(NO3)3
Step 1: Balance the equation, which in this case the equation
doesn’t need balancing because it already is.
Step 2: Identify what the question asks and what you have been
given. We are asked to find the amount of moles needed in
order to react with 12.3 moles of nitric acid. 12.3 moles of nitric
acid is our given information that we have to use to solve fore
aluminum hydroxide.
Step 3: Set up the equation with given information. We must use
12.3 moles of nitric acid to solve.
Step 4: Solve. This is what the equation should look like.
12.3 mol HNO3 x 1 mol Al(OH)3 (We must divide by 3 mol HNO3 to get
3 mol HNO3
rid of nitric acid)
12.3 x 1/3 = 4.1 mol Al(OH)3 This is the final answer.
Problem #2
How many grams of carbon dioxide are produced when 46.8 grams of
C2H2 combust?
2 C2H2 + 5 O2 → 4 CO2 + 2 H2O
Step 1: Balance the equation, which it does not need balancing.
Step 2: Identify what you are asked to find and what you have been
given(46.8 g of C2H2).
Step 3: Set up an equation with 46.8 g of C2H2 as your starting amount.
Step 4: Solve for carbon dioxide. The equation looks like this.
46.8 g C2H2 x 1 mol C2H2 x 4 mol CO2 x 44.01 g CO2 = 158. 2 g CO2
26.03g C2H2 2 mol C2H2 1 mol CO2
• If you are wondering where the grams came from in the equation, they
are the atomic mass of elements combined with other elements into one
mass. We come up with the final answer bymultiplying everything in the
numerator and dividing by everything in the denominator. 158.2 grams
of carbon dioxide is the final answer when we have multiplied and
divide everything.
Problem#3: examples, examples and more examples
How many grams of chlorine gas can be produced when
62.3 grams of aluminum chloride decompose?
2 AlCl3 →2 Al + 3 Cl2
Step 1: Balance the equation.
Step 2: Identify your given info and what you are asked to
find(grams of chlorine gas and 62.3 grams of aluminum
chloride).
Step 3: Set up an equation with your given information.
Step 4: Solve for grams of chlorine gas.
62.3 g AlCl3 x 1 mol AlCl3 x 3 mol Cl2 x 70.8g Cl2 = 49.7 g
133.3 g AlCl3 2 mol AlCl3 1 mol Cl2 Cl2 is
the final answer