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# STOICHIOMETRY

A thorough look into the true chemistry behind stoichiometry.
by

## lauren widaman

on 26 August 2014

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#### Transcript of STOICHIOMETRY

STOICHIOMETRY
TELL ME SOME MOLE...
PERCENT YIELD
Stoichiometry versions
the basics
The simplest conversions are those that go from mol to mol. To convert from mol to mol, we need to use the
mole ratio
derived from the coefficients within the balanced reaction equation.
Limiting reactant/Reagent
This is a stoich problem with no available quantities of reactants in the EXACT molar ratios. There are one or more reactants which are "in excess" and the other(s) is the "limiting reactant(s)." You can recognize these types of problems when given a mass, volume or mols of more than one reactant. To determine the ACTUAL amount of product(s) produces, you first have to figure out which reactant "limits" the reaction.

ARE YOU SMARTER THAN SWANGO?
Yoon Cho, Emily Liu, Amy Wang and Lauren Widaman
moles
particles
volume at STP
grams
÷
÷
÷
x
x
x
6.022 x 10^23
molar mass (mm)
22.4 L
Because substances contain so many atoms, a unit of measure called the
mole
was created to count atoms. The mole,
abbreviated as mol
, is the number of carbon atoms in 12 grams of pure Carbon-12, therefore determining a mole equal to
6.022*10^23
. This number is called
.
mole ratio =
mol B
mol A
4NH + 5O
How many mols of O will we get if we have 2.8 mol NH ?
3 2 2
4NO + 6H O
2 3
2.8 mol NH ( ) =
5 mol O
3
2
4 mol NH
3
#1
4Al + 3O 2Al O
#2
2Mg + O 2MgO
#3
2KClO 2KCl + 3O
#4
As much as you try, you can't get same answer in your calculations as you get in the real experiment. The pre-calculated answer is always a
theoretical value
, a value that can rarely be achieved because of other various factors that are in play during the experiment. Whether it's personal, systematic, or random error- even if you think your answer is really close- it's still not the 'perfect' answer. So to calculate how accurate your answer is, we use
percent yield
, and it goes a lil' somethin' like this:
Calculated result: 53 grams

How to Percent Yield
x
100 = %YIELD
51.3
53
x
100 =
1
2
0.97 x 100 =
Percent Yield = 97%
3
le example:
2 2 3
If there are 0.25 mol Al and 0.40 mol of O involved in a reaction, what is the limiting reactant?
2
2Na O(s) + 3CO(g) Na CO (s) +2Na(s)
2 2 3
What mass (in grams) of Na CO is produced if 50.0g of Na O react with
excess CO?
2 3 2
2
Burning 20.0g of Mg in the presence of O produces 32.5g of the product. What is the percent yield?
2
the baby steps
1. Write the balanced chemical reaction equation.
2. Convert each of the reactants into the same product.
3. Compare the results from Step #2. The reactant that produces the least amount of product is the
limiting reactant
(as you can guess... the other is
in excess
, also known as the
excess reactant
).
use YOUR BABY steps and start walking!
What is the limiting reactant when 2.5 mols of hydrogen react with 1.5 mols of oxygen to produce water?
Step UNO
2H + O
Write the balanced chemical reaction equation.
2 2
2H O
2
20g Mg (1mol Mg) (2mol MgO) (40g MgO)
(24g Mg) (2mol Mg) (1mol MgO)
32.5 / 33.3 x 100 = 97.9%

50g Na O (1mol Na O) (1mol Na CO ) (106g Na CO )
(62g Na O) (2mols Na O) (1mol Na CO )
2 2 2 3
2 2 2 3 2 3
Step dos
2H + O 2H O

Al

0.25 Al (2 mol Al O /4 mol Al)=0.125 mol

0.40 O (2mol Al O /3mol O )= 0.267 mol
2 3
2 2 3 2
Convert each of the reactants into the same product.
3 2
How many liters of O come from the decomposition of 850g of KClO ?
(the density of O is 1.429 g/L.)
2
3
2
850g KClO (1mol KClO ) (3mol O ) (32g O ) ( 1L O )
3 3 2 2 2
(122g KClO ) (2mol KClO ) (1mol O ) (1.429g O )
3 3 2 2
In stoichiometry, mol will always be a connecting conversion, if not one of the two targeted units.
For example, as shown earlier, we may have mole to mole calculations. Along with that, we may be asked to convert from grams to moles, particles to volume, etc.
2 2 2
----
----
2.5 mol
1.5 mol
2.5 mol H
( )
2
2 mol H O
2 mol H
2
2
= 2.5 mol H O
2
MOLE RATIO
1.5 mol O
2
( )
2 mol H O
2
1 mol O
2
= 3.0 mol H O
2
Step tres
2.5 mol H O
<<<<<<<
3.0 mol H O
Compare the results!
2 2
Therefore, we can conclude that
H
is
the
limiting reactant
because it produces less product than
H O
.
2
2
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