Mr. Anthony Gates
Chemical Equations:
The symbolic representation of a chemical reaction.
Skeleton Equations:
Chemical equation that tells us what is involved,
but not the relative amounts.
Balanced Equation:
Chemical equation that tells us what is involved,
and the relative amounts.
NET IONIC EQUATIONS!!!
But first…
Imagine that you are an alien sent to Earth to
learn about who can be found at Baseball
stadiums.
Baseball players=Ions that are actively involved
in the reaction.
Crowd in the stands=spectator ions
Net ionic equations ignore spectator ions and
focus on any ions that are involved in the
reaction.
Net ionic equations ignore the fans in the
stadium and focus on the players actively
playing the game.
Balanced Equation:
𝑃𝑏 𝑠 + 2𝐴𝑔𝑁𝑂3 𝑎𝑞 → 2𝐴𝑔 𝑠 + 𝑃𝑏(𝑁𝑂3 )2 𝑎𝑞
Complete Ionic Equation:
𝑃𝑏 𝑠 + 2𝐴𝑔+ 𝑎𝑞 + 2𝑁𝑂3 − 𝑎𝑞 → 2𝐴𝑔 𝑠 + 𝑃𝑏2+ (𝑎𝑞) + 2𝑁𝑂3 − 𝑎𝑞
𝑁𝑂3 is a spectator ion… a fan at the baseball stadium.
Balanced Net Ionic Equation:
𝑃𝑏 𝑠 + 2𝐴𝑔+ 𝑎𝑞 → 2𝐴𝑔 𝑠 + 𝑃𝑏 2+ 𝑎𝑞
1. Write a balanced equation with phase symbols.
2. Separate any aqueous compounds.
• Make sure to keep this balanced by distributing
coefficients.
3. Write in charges for any lone ions.
4. Cross out anything that is the same on both
sides of the equation.
• These are the spectator ions… the fans in the
baseball stadium.
5. Put what is left back into a balanced equation.
Chemical Equation:
2 NaOH(aq) + CuCl2(aq)  2 NaCl(aq) + Cu(OH)2(s)
Complete Ionic Equation:
(all ions in aqueous solution are broken apart and gases,
precipitates, and water are left unchanged)
2 Na+(aq) + 2 OH-(aq) + Cu+2(aq) + 2 Cl- (aq)  2 Na+(aq) + 2 Cl-(aq)
+ Cu(OH)2(s)
Spectator Ions: Na+ and Cl-
Net Ionic Equation:
Cu+2(aq) + 2 OH-(aq)  Cu(OH)2(s)
Some double replacement reactions produce
compounds that immediately react to undergo another
reaction.
Overall Ionic Equations are used to show the net ionic
equations from the very start to the very end as one
reaction.
Simply combine the reactions to find the Overall Ionic
Equation.
Ex: RXN 1: HCl(aq) + NaHCO3(aq)  H2CO3(aq) + NaCl(aq)
RXN 2: H2CO3 (aq)  H2O (l) + CO2 (g)
RXN 1: HCl(aq) + NaHCO3(aq)  H2CO3(aq) + NaCl(aq)
RXN 2: H2CO3 (aq)  H2O (l) + CO2 (g)
Combined equations:
HCl(aq) + NaHCO3(aq) + H2CO3 (aq)  H2CO3(aq) +NaCl(aq) + H2O(l) + CO2(g)
Complete Ionic Equation:
H+(aq) + Cl-(aq) + Na+(aq) + HCO3-(aq)  Na+(aq) + Cl-(aq) + H2O(l) + CO2(g)
Net Ionic Equation:
H+(aq) + HCO3-(aq)  H2O(l) + CO2(g)
How many molecules are in 720g of
𝐶6 𝐻12 06 (𝑠)?
720g 𝑪𝟔 𝑯𝟏𝟐 𝟎𝟔
1mol 𝑪𝟔 𝑯𝟏𝟐 𝟎𝟔
𝟔. 𝟎𝟐𝒙𝟏𝟎𝟐𝟑 molecules
180.0g 𝑪𝟔 𝑯𝟏𝟐 𝟎𝟔 1mol 𝑪𝟔 𝑯𝟏𝟐 𝟎𝟔
= 2.4𝑥1024 molecules𝐶6 𝐻12 06
What is the mass of 32.0L of 𝐶𝑙2 at STP?
32.0L 𝑪𝒍𝟐
1mol 𝑪𝒍𝟐
70.8g 𝑪𝒍𝟐
22.4L 𝑪𝒍𝟐
1mol 𝑪𝒍𝟐
=101g 𝐶𝑙2
How much space will 6.38 x 1023 molecules of
NH3 take up under standard temperature and
pressure?
How many molecules are in 157.9g of NaCl at
STP?
What is the mass of 58.9L of CO2 at STP?
How many molecules are in 720g of C6H12O6?
What is the volume of 57.9g of methane CH4
gas at STP?
What is the mass of 1.5 x 1025 molecules of
CO2?
Because following a recipe needed a big word
that was hard to say
Stoichiometry is the calculation of how much of
a product will be produced based on how much
of the reactants you used, or vice versa.
Say a sandwich comes with: 1 bun, 2 slices of
ham, 3 slices of turkey and 1 packet of mayo. An
equation could be written as:
1B+2H+3T+1M1S
1B+2H+3T+1M1S
For 1 sandwich you need 2 slices of ham. So for
2 sandwiches, you need 4 slices of ham.
This is a mole ratio.
2H:1S
We can use these ratios to determine how
much of a reactant will be needed to make
some amount “x” of product
How much ham would you need to make a
dozen (12) sandwiches?
1B+2H+3T+1M1S
12 sandwiches
2 slices of ham
1 sandwich
= 24 𝑠𝑙𝑖𝑐𝑒𝑠 𝑜𝑓 ℎ𝑎𝑚
2𝐻2 𝑂2 (𝑙) → 2𝐻2 𝑂(𝑙) + 𝑂2 (𝑔)
How much oxygen gas would created from
decomposing 8mol of hydrogen peroxide?
8mol 𝑯𝟐 𝑶𝟐 (l)
1mol 𝑶𝟐 (g)
2mol 𝑯𝟐 𝑶𝟐 (l)
= 4𝑚𝑜𝑙 𝑂2 (𝑔)
2𝐶2 𝐻6 (𝑔) + 7𝑂2 (𝑔) → 6𝐻2 𝑂(𝑙) + 4𝐶𝑂2 (𝑔)
How much dicarbon hexahydrogen (𝐶2 𝐻6 ) is
needed to produce 28mol of water?
28mol 𝑯𝟐 𝑶(𝒍)
2mol 𝑪𝟐 𝑯𝟔 (𝒈)
6mol 𝑯𝟐 𝑶(𝒍)
= 9. 3𝑚𝑜𝑙𝐶2 𝐻6 (𝑔)
Mole-to-Mole ratios
Another form of dimensional analysis comparing
the amount of moles of one substance to the
amount of moles of another substance.
Use the coefficients of the balanced equation.
With a good understanding of mole-to-mole
ratios and mixed mole conversions, the world is
your oyster.
With these concepts we can calculate how
much products will be made based on how
much of the reactant we are starting with, or
vice versa.
Let’s start with an easy equation…
2𝐶2 𝐻6 (𝑔) + 7𝑂2 (𝑔) → 6𝐻2 𝑂(𝑙) + 4𝐶𝑂2 (𝑔)
What is the mass of the carbon dioxide that is
produced if 97.4g of 𝑂2 𝑔 reacts?
Well, let’s break up this problem into parts…
How many moles make up 97.4g of 𝑂2 𝑔 ?
How many moles of carbon dioxide are produced?
What is the mass of carbon dioxide that is
produced?
2𝐶2 𝐻6 (𝑔) + 7𝑂2 (𝑔) → 6𝐻2 𝑂(𝑙) + 4𝐶𝑂2 (𝑔)
What is the mass of the carbon dioxide that is
produced if 97.4g of 𝑂2 𝑔 reacts?
How many moles make up 97.4g of 𝑂2 (𝑔)?
97.4g 𝑂2
1mol 𝑂2
32.0g 𝑂2
= 3.04375mol 𝑂2
2𝐶2 𝐻6 (𝑔) + 7𝑂2 (𝑔) → 6𝐻2 𝑂(𝑙) + 4𝐶𝑂2 (𝑔)
What is the mass of the carbon dioxide that is
produced if 97.4g of 𝑂2 𝑔 reacts?
How many moles of carbon dioxide are produced?
3.04375mol 𝑂2
4 mol 𝐶𝑂2
7mol 𝑂2
=1.73928mol 𝐶𝑂2
2𝐶2 𝐻6 (𝑔) + 7𝑂2 (𝑔) → 6𝐻2 𝑂(𝑙) + 4𝐶𝑂2 (𝑔)
What is the mass of the carbon dioxide that is
produced if 97.4g of 𝑂2 𝑔 reacts?
What is the mass of carbon dioxide that is
produced?
1.73928mol 𝐶𝑂2
44.0g 𝐶𝑂2
1 mol 𝐶𝑂2
=76.5g 𝐶𝑂2
2𝐶2 𝐻6 (𝑔) + 7𝑂2 (𝑔) → 6𝐻2 𝑂(𝑙) + 4𝐶𝑂2 (𝑔)
What is the mass of the carbon dioxide that is
produced if 97.4g of 𝑂2 𝑔 reacts?
97.4g 𝑂2
1mol 𝑂2
4 mol 𝐶𝑂2 44.0g 𝐶𝑂2
32.0g 𝑂2
7mol 𝑂2
1 mol 𝐶𝑂2
=76.5g 𝐶𝑂2
2𝐶2 𝐻6 (𝑔) + 7𝑂2 (𝑔) → 6𝐻2 𝑂(𝑙) + 4𝐶𝑂2 (𝑔)
What is the mass of the carbon dioxide that is
produced if 97.4g reacts?
2 H2 (g) + O2 (g) → 2 H2O (g)
How many molecules of O2 gas are needed to
produce 25.4 H2O?
When does a chemical reaction stop?
When the reactants are used up… When all of
the reactants are used up?
When one of the reactants are used up.
As soon as one of the reactants runs out, the
reaction stops.
Mr. Gates wants to make S’mores for his students. He
has 100 students total and so he needs a lot of
ingredients.
By the end of the day the other teachers found out
that Mr. Gates’ students made S’mores that day and
they asked if they could make some with the leftovers.
He of course said yes, but wasn’t sure how many could
still be made.
Mr. Gates still has half of box of graham crackers (with
15 graham crackers inside), 6 fun-size chocolate bars
(each of which can easily be broken up into 4 pieces),
and a bag of 9 marshmallows.
What is the most amount of S’mores that could
be made with the leftovers if the balanced
chemical equation for a S’more is:
2𝐺𝑐 + 1𝑀 + 4𝐶𝑝 → 1𝑆𝑚
Leftovers:
15Gc (graham crackers)
9M (marshmallows)
24Cp (chocolate pieces)
What is the most amount of S’mores that could
be made with the leftovers if the balanced
chemical equation for a S’more is:
2𝐺𝑐 + 1𝑀 + 4𝐶𝑝 → 1𝑆𝑚
15Gc
1Sm
2Gc
=7.5 Sm
So we can make 7 S’mores with the graham
crackers available.
What is the most amount of S’mores that could
be made with the leftovers if the balanced
chemical equation for a S’more is:
2𝐺𝑐 + 1𝑀 + 4𝐶𝑝 → 1𝑆𝑚
9M
1Sm
1M
= 9𝑆𝑚
So we can make 9 S’mores with the
marshmallows available.
What is the most amount of S’mores that could
be made with the leftovers if the balanced
chemical equation for a S’more is:
2𝐺𝑐 + 1𝑀 + 4𝐶𝑝 → 1𝑆𝑚
24Cp
1Sm
4Cp
= 6𝑆𝑚
So we can make 6 S’mores with the chocolate
pieces available.
Gc… 7 S’mores
M… 9 S’mores
Cp… 6 S’mores
We can only make 6 S’mores before we run out of
an ingredient. The ingredient we run out of is
Chocolate (isn’t always)
This is what we call a limiting reagent.
Limiting reagents are reactants that limit the
amount of product that can be produced.
All of the reagent will be used up in the reaction.
Excess reagent is a reactant that we have more
of than we need.
There will be extra of this ingredient that didn’t
react when the reaction is done.
Anything that is not a limiting reagent is an excess
reagent.
Copper reacts with sulfur to form copper (I)
sulfide according to the following balanced
equation:
2𝐶𝑢 𝑠 + 𝑆(𝑠) → 𝐶𝑢2 𝑆(𝑠)
A)What is the limiting reagent when 80.0g Cu
reacts with 25.0g S?
B)Amount of product being produced?
C)How much excess reagent is left over?
What is the limiting reagent when 80.0g Cu reacts
with 25.0g S?
2𝐶𝑢 𝑠 + 𝑆(𝑠) → 𝐶𝑢2 𝑆(𝑠)
80.0gCu
1mol Cu
63.5gCu
25.0gS
1molS
32.1gS
1.26molCu
1molS
2molCu
= 1.26𝑚𝑜𝑙𝐶𝑢
= 0.779𝑚𝑜𝑙𝑆
= .630𝑚𝑜𝑙𝑆
Amount of product produced when 80.0g Cu
reacts with 25.0g S?
2𝐶𝑢 𝑠 + 𝑆 𝑠 → 𝐶𝑢2 𝑆 𝑠
1.26molCu
1mol𝑪𝒖𝟐 𝑺
159.1g𝑪𝒖𝟐 𝑺
2molCu
1mol𝑪𝒖𝟐 𝑺
= 100. 𝑔𝐶𝑢2 𝑆
How much excess reagent is left over when
80.0g Cu reacts with 25.0g S?
2𝐶𝑢 𝑠 + 𝑆(𝑠) → 𝐶𝑢2 𝑆(𝑠)
0.779𝑚𝑜𝑙𝑆 − .630𝑚𝑜𝑙𝑆 = .149𝑚𝑜𝑙𝑆
0.149molS
32.1gS
1molS
= 4.78𝑔𝑆
Ex: What is the limiting reactant if 8.0 mol of HF
reacts with 4.5 mol of SiO2?
SiO2 + 4 HF  SiF4 + 2 H2O
a) Limiting Reactant?
b) Amount of Products being produced?
c) How much of the Excess Reactant is left over?
Ex: What is the limiting reactant if 8.0 mol of HF
reacts with 4.5 mol of SiO2?
SiO2 + 4 HF  SiF4 + 2 H2O
Limiting Reactant?
8.0 mol HF 2 mol H2O = 4.0 mol H2O
4 mol HF
4.5 mol SiO2 2 mol H2O
1 mol SiO2
= 9.0 mol H2O
Ex: What is the limiting reactant if 8.0 mol of HF
reacts with 4.5 mol of SiO2?
SiO2 + 4 HF  SiF4 + 2 H2O
Limiting Reactant?
8.0 mol HF 1 mol SiO2 = 2.0 mol SiO2
4 mol HF
Do we have enough SiO2 for all of the HF to
react?
YES, so all of the HF will react (LR)
Ex: What is the limiting reactant if 8.0 mol of HF
reacts with 4.5 mol of SiO2?
SiO2 + 4 HF  SiF4 + 2 H2O
Limiting Reactant?
4.5 mol SiO2 4 mol HF
= 18 mol HF
1 mol SiO2
Do we have enough HF for all of the SiO2 to
react?
NO, HF is the limiting reactant
Ex: What is the limiting reactant if 8.0 mol of HF
reacts with 4.5 mol of SiO2?
SiO2 + 4 HF  SiF4 + 2 H2O
Amount of Products Produced?
8.0 mol HF 2 mol H2O = 4.0 mol H2O
4 mol HF
8.0 mol HF 1 mol SiF4
4 mol HF
= 2.0 mol SiF4
Ex: What is the limiting reactant if 8.0 mol of HF
reacts with 4.5 mol of SiO2?
SiO2 + 4 HF  SiF4 + 2 H2O
How much of the excess reactant is left over?
8.0 mol HF 1 mol SiO2 = 2.0 mol SiO2
4 mol HF
4.5 mol SiO2 - 2.0 mol SiO2 = 2.5 mol SiO2
2Al + 6HCl  2AlCl3 + 3H2
If 25.0 g of aluminum was added to 90.0 g of
HCl, what mass of H2 will be produced?
Hint: first figure out who is limiting, then use
stoich to calculate the product.
Answer:
2.47g Hydrogen diatomic
N2 + 3H2  2NH3
If you have 20.0 g of N2 and 5.00 g of H2, which
is the limiting reagent?
Answer:
Nitrogen diatomic
4 Al + 3 O2  2Al2O3
What mass of aluminum oxide is formed when
10.0 g of Al is burned in 20.0 g of O2?
Answer:
18.9gAl2O3
C3H8 + 5 O2  3 CO2 + 4 H2O
If 15.0 g of C3H8 reacts with 60.0 g of O2, how
many grams of CO2 are produced?
Answer
45.0gCO2
C3H8 + 5 O2  3 CO2 + 4 H2O
If 15.0 g of C3H8 reacts with 60.0 g of O2, how
much of the excess is left unreacted?
Answer: 0.17𝑚𝑜𝑙𝑂2
Stoich: We only know the amount of one
substance and want to find the amount of
another.
Limiting: we know the amount of multiple
substances and want to find the amount of
another.
This is the maximum amount of product that
can be made in a chemical reaction.
This must be calculated using stoichiometry.
This is the measured amount of product that is
produced in a lab setting.
Almost always less than theoretical yield
Reactants may form unexpected products
Purification of the product usually results in losing some product
Some of the product may have been lost during the lab
If the Actual Yield is more than the Theo. Yield, you
have some substance acting as a false product
This is the relationship between theoretical and
actual yields.
The closer to 100%, the better the lab was
performed
% Yield =
Actual Yield
Theoretical Yield
x 100
***units for Act. Yield and Theo. Yield must be the same
In lab 38.8 g of C6H5Cl are collected. If you
started with 36.8 g of C6H6 and excess Cl2, what
is the percent yield?
C6H6 + Cl2  C6H5Cl + HCl
36.8 g C6H6
1 mol C6H6
78.0 g C6H6
1 mol C6H5Cl
1 mol C6H6
38.8 g C6H5Cl x 100 = 73.1 %
53.1 g C6H5Cl
112.5 g C6H5Cl
1 mol C6H5Cl
= 53.1 g C6H5Cl
𝐶𝑢𝐶𝑙2 + 2𝑁𝑎𝑁𝑂3 → Cu(𝑁𝑂3 )2 + 2NaCl
If 15 grams of copper (II) chloride react with 20.
grams of sodium nitrate, how much sodium
chloride can be formed?
What is the name of the limiting reagent?
DO NOT ERASE YOUR ANSWERS!!
𝐶𝑢𝐶𝑙2 + 2𝑁𝑎𝑁𝑂3 → Cu(𝑁𝑂3 )2 + 2NaCl
How much of the excess reagent is left over in
this reaction?
If 11.3 grams of sodium chloride are formed in
the reaction, what is the percent yield of this
reaction?
3𝐶𝑎𝐶𝑂3 + 2𝐹𝑒𝑃𝑂4 →𝐶𝑎3 (𝑃𝑂4 )2 + 𝐹𝑒2 (𝐶𝑂3 )3
Which reactant is limiting, assuming we start
with 100. grams of calcium carbonate and 45.0
grams of iron (III) phosphate.
What is the mass of each product that can be
formed?
What mass of the excess reactant(s) is left over?