Chapter 12 Stoichiometry1.ppt-chemistry.

chemistry

1. Mathematics of Mathematics of
Chemical EquationsChemical Equations
By using “mole to mole” By using “mole to mole”
conversions and conversions and balancedbalanced
equationsequations, we can calculate the , we can calculate the
exact amounts of substances exact amounts of substances
that will be used up or produced that will be used up or produced
during a reaction.during a reaction.

2. •What you’ll learn:
•Write mole ratios from a balanced chemical
equation
•Calculate the number of moles and the mass of a
reactant or product when given the number of
moles or the mass another reactant.
•Identify the limiting reactant in a chemical
reaction.
•Determine the percent yield of a chemical reaction.

3. Basic ExampleBasic Example
NN
22 + 3H + 3H
22  2NH 2NH
33
•From this balanced equationFrom this balanced equation, we know that for , we know that for
every 1 mole of Nevery 1 mole of N
22 we need ___ moles of H we need ___ moles of H
22 to to
react in order for ___ moles NHreact in order for ___ moles NH
3 3 to be producedto be produced
..
•How many moles of NHow many moles of N
22 would we need to react would we need to react
with 6 moles of Hwith 6 moles of H
22??
6 moles
H
2 moles H
23
moles
N
2
1
= 2 moles of N
2
x
3
2

4. NN
22 + 3H + 3H
22  2NH 2NH
33
•How many moles of Ammonia (NHHow many moles of Ammonia (NH
33) are ) are
produced if you have an excess of produced if you have an excess of
Nitrogen gas, and a sample of 0.43 Nitrogen gas, and a sample of 0.43
moles of Hydrogen?moles of Hydrogen?
0.43 moles H
2
moles H
23
moles NH
32
= 0.287 moles
of NH
3
x

5. Mass – Mass ConversionsMass – Mass Conversions
•You MUST BALANCE the equation first!You MUST BALANCE the equation first!
NaClONaClO
33  NaCl + O NaCl + O
22
How many grams of OHow many grams of O
22 would be produced if would be produced if
the final mass of NaCl produced was 375.6 g?the final mass of NaCl produced was 375.6 g?
375.6 g NaCl375.6 g NaCl

2 2 3
x
58.5 g NaCl
mole NaCl1
x x
mole NaCl
mole O
2
2
3
mole O
2
1
32 Grams O
2
= 308.18g O
2

6. Stoichiometric Conversions Stoichiometric Conversions
A Three Step ProcessA Three Step Process
•Step 1 Convert the value of the given Step 1 Convert the value of the given
substance to moles of that substance.substance to moles of that substance.
•Step 2 Convert from moles of the given Step 2 Convert from moles of the given
substance to moles of the unknown substance to moles of the unknown
substance (use the mole ratio from the substance (use the mole ratio from the
balanced equation).balanced equation).
•Step 3 Convert from moles of the Step 3 Convert from moles of the
unknown to proper unit. unknown to proper unit.

7. Limiting ReactantsLimiting Reactants

8. Limiting ReactantsLimiting Reactants
•When calculating your theoretical When calculating your theoretical
products, it is important to factor in products, it is important to factor in
which reactant you will run out of firstwhich reactant you will run out of first! !
•This is called the “Limiting Reactant”. This is called the “Limiting Reactant”.
The other reactant(s) is called the The other reactant(s) is called the
“Excess Reactant”.“Excess Reactant”.

9. Ex:Ex: If you have 10grams of OIf you have 10grams of O
2 2 and 10 Liters of and 10 Liters of
HH
22, how many molecules of water can you , how many molecules of water can you
create?create?
2H2H
22 + O + O
22  2H 2H
22OO
•Step 1: Convert both to moles.Step 1: Convert both to moles.
10gO10gO
2 x 2 x 1 mole O 1 mole O
22 = = 0.31 moles O0.31 moles O
22
1 32g O1 32g O
22
10L 10L HH
2 x 2 x 1 mole H 1 mole H
2 = 2 = 0.45 moles H0.45 moles H
22
1 22.4L H1 22.4L H
22

10. •Step 2: Complete a mole to mole Step 2: Complete a mole to mole
conversion, from one reactant to the other conversion, from one reactant to the other
reactant to find the theoretical need.reactant to find the theoretical need.
2H2H
22 + O + O
22  2H 2H
22OO
0.31 moles O0.31 moles O
22 x x 2 moles H2 moles H
22
= = 0.62 moles H0.62 moles H
22
1 mole O1 mole O
2 2 NEEDEDNEEDED
YOU KNOWYOU KNOW that you have 0.45 moles H that you have 0.45 moles H
22
from your given values. So, that means from your given values. So, that means
you do not have enough Hyou do not have enough H
22. It is therefore . It is therefore
the Limiting Reactant.the Limiting Reactant.

11. •IF YOU COMPLETED STEP 2 FINDING IF YOU COMPLETED STEP 2 FINDING
MOLES OF OMOLES OF O
22::
2H2H
22 + O + O
22  2H 2H
22OO
0.45 moles H0.45 moles H
22 x x 1 moles O1 moles O
22
= = 0.225 moles O0.225 moles O
22
2 mole H2 mole H
2 2 NEEDEDNEEDED
You would come to the same conclusion, You would come to the same conclusion,
because this shows that you have more because this shows that you have more
than enough Othan enough O
22. (You have .31 moles). (You have .31 moles)

12. •Step 3: Decide from Step 2 which Step 3: Decide from Step 2 which
reactant is the L.R. and use its moles to reactant is the L.R. and use its moles to
calculate your answer. (MUST have a calculate your answer. (MUST have a
balanced Chemical Equation!)balanced Chemical Equation!)
2H2H
22 + O + O
22  2H 2H
22OO
•0.45mol H0.45mol H
22 x x 2 mol H2 mol H
22OO
x x 6.02x106.02x10
2323
molecules H molecules H
22OO
= =
11 2 mol H 2 mol H
22 1 mole H1 mole H
22OO
2.71x102.71x10
2323
molecules H molecules H
22OO

13. Percent YieldPercent Yield
•Most chemical reactions never succeed in Most chemical reactions never succeed in
producing the predicted amount of producing the predicted amount of
product. product.
•So, the actual amount of product is less So, the actual amount of product is less
than expected, due to experimental error than expected, due to experimental error
•This is generally due to experimental This is generally due to experimental
error such as evaporation, product left error such as evaporation, product left
on filter paper etc.on filter paper etc.

14. Percent YieldPercent Yield
•Percent yield is the ratio of actual yield Percent yield is the ratio of actual yield
to the theoretical yield expressed as a to the theoretical yield expressed as a
percent. percent.