# Mol ? Mol Stoichiometry by S0U2D663

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```									           Stoichiometry
Compare the amounts of the different
substances in a chemical reaction.
Can measure amount in two ways: moles or
mass.
The balanced equation gives the ratio of
moles involved in the reaction.
Take a reaction:
2Na + CaSO4  Na2SO4 + Ca
What does this mean?
2 sodium atoms reacts with 1 formula unit of
CaSO4 to make 1 formula unit of Na2SO4
and 1 calcium atom.
Na. This ratio must be true.
Thus: 1 mol of CaSO4 will react with 2 mol of
Na to produce 1 mol of Na2SO4 and 1 mol
of Ca.
We call these mole ratios and allow us to
relate the amount of the substances
involved in the reaction
Mol  Mol stoichiometry
How many moles of Na do we need to react
with the CaSO4?
Equation: 2Na + CaSO4  Na2SO4 + Ca
5.00 mol   2.50 mol    2.50 mol    2.50 mol

Write what you are given:           Each has a 1/1 ratio
2.5 mol CaSO4          2 mol Na    = 5.00 mol Na

1 mol CaSO4
How many moles of each product?


Mol  Mol stoichiometry
To set up the ratio, you put the substance
you want to solve for on the top and the
one you are given on the bottom.
Stoichiometry with mass
to moles first using your molar mass
• Then you use your mole ratio
• If you want to end with mass you use your
mole ratio
• Then you convert to mass using your
molar mass
Problems with Mass
Example 1: 0.125 mol of calcium chloride reacts
with sodium phosphate.
a) What mass of the sodium phosphate is needed
in the reaction?
b) What mass of each product is created?
Solution
First write the balance equation
3 CaCl2 + 2 Na3PO4        Ca3(PO4)2 + 6 NaCl

a) 0.125 mol CaCl2 * 2 mol Na3 PO4   163.940 g Na3PO4
                  
3 mol CaCl2     1 mol Na3PO4

= 13.7 g Na3PO4
              
Solution
3 CaCl2 + 2 Na3PO4             Ca3(PO4)2 + 6 NaCl
b) 0.125 mol CaCl2 * 1 mol Ca3 (PO4 ) 2     310.174 g Ca3 (PO4 )2
                       
3 mol CaCl2         1 mol Ca3 (PO4 )2
= 12.9 g Ca3(PO4)2


0.125 mol CaCl2 *     6 mol NaCl      58.443 g NaCl
                   
3 mol CaCl2        1 mol NaCl
= 14.6 g NaCl

              
Problems with mass
Example 2: 2.50 g of calcium reacts with
silver nitrate.
a) How many mol of silver nitrate are
needed?
b) How many mol of each product are
created?
Solution
First write the balance equation
Ca + 2 AgNO3  Ca (NO3)2 + 2 Ag
0.0624 mol   0.125 mol

a) 2.50 g Ca *    1 mol Ca       2 mol AgNO3
               
40.078 g Ca       1 mol Ca

= 0.125 mol AgNO3

              
Mass  Mass Stoichiometry
Example 1: 6.25 g of iron (II) sulfate reacts
with Aluminum
a) What mass of Al is needed for the
reaction?
b) What is the theoretical yield of each
product?
First write the balanced equation:
2Al + 3FeSO4  Al2(SO4)3 + 3Fe
Solution part a:

1 mol FeSO4     2 mol Al      26.982 g Al
6.25 g FeSO4                             
151.906 g    3 mol FeSO4      1 mol Al

= 0.740 g

First convert 
from       Use mole
       Convert to
mass to moles                             mass
ratio
Use molar
Use molar mass
mass
Solution part b: first solve for Al2(SO4)3
1 mol FeSO4   1 mol Al2 (SO4 ) 3      342.147 g
6.25 g FeSO4                                  
151.906 g      3 mol FeSO4        1 mol Al2 (SO4 ) 3

= 4.69 g
Note: the first part of the
equation is thesame –                 
replace with [ ]
Solve for Fe

[ ]           3 mol Fe           55.845 g         = 2.30 g
                    
3 mol FeSO4          1 mol Fe

                   
Mass  Mass Stoichiometry
Note: if you add the mass of the products
together it should equal the combined
mass of reactants

Reactants: 6.25 g + 0.740 g = 6.99 g
Products: 4.69 + 2.30 = 6.99 g
Example 2:
15.3 g of calcium reacts with silver nitrate
a) What mass of silver nitrate is needed for
the reaction?
b) What is the theoretical yield of each
product?
Balanced equation:
Ca + 2AgNO3  Ca(NO3)2 + 2Ag

Masses:
AgNO3 = 130. g
Ca(NO3)2 = 62.6 g
Ag       = 82.4 g
Work for example 2
15.3 g Ca * 1 mol Ca * 2 mol AgNO3 * 169.874 g AgNO3 =
40.078 g Ca   1 mol Ca      1 mol AgNO3
130. g AgNO3
[ ] * 1 mol Ca(NO3)2 * 164.086 g Ca(NO3)2 = 62.6 g Ca(NO3)2
1 mol Ca          1 mol Ca(NO3)2

[ ] * 2 mol Ag * 107.87 g Ag = 82.4 g Ag
1 mol Ca    1 mol Ag
Limiting reactants
In real reactions one of the reactants will
always run out first. At this point the
reaction stops.
The amount of this reactant limits the
amount of product created and is called
the limiting reactant.
There will be some of the other reactant left
over….thus it is present in excess.
Limiting reactants example
Reaction:         C (s) + O2 (g)  CO2 (g)

Say you mix together 5 mol of C and 10 mol of O2?
What will happen?
All 5 mol of the C will react at which point the
reaction will stop. Thus the C is the limiting
reactant.
Based on our stoichiometry: all the C will react with
5 mol of O2 to produce 5 mol of CO2.
How much excess O2 will be left over?
Start with 10 mol, 5 mol reacts so 5 mol will be left
over.
Example #2
2.0 mol of HF reacts with 4.5 mol of SiO2.
The reaction is: SiO2 + 4HF  SiF4 + 2H2O
Which is the limiting reactant?
Pick one of the reactants: Say we pick SiO2.
Use Stoichiometry to figure out how much of the
other reactant is needed:
4.5 mol SiO2 * 4 mol HF = 18 mol HF needed
1 mol SiO2
But you only have 2.0 mol of HF. Thus the HF is
limiting!!!!
Example #2 continued
What if we had picked the HF?
2.0 mol HF * 1 mol SiO2 = 0.50 mol SiO2
4 mol HF
There is 4.5 mol of SiO2 so there is more
than enough SiO2. So it is in excess and
the HF is limiting.
So how much excess SiO2?
4.5 mol – 0.50 mol = 4.0 mol excess SiO2
Yet more example #2
How many mol of each product will be
created?
The stoichiometry is based on the limiting
reactant which is the HF, so:
SiO2 + 4HF  SiF4 + 2H2O
0.50     2.0     0.50 1.0
mol          mol      mol mol
Example #3
Same reaction: 3.5 mol of HF reacts with
0.50 mol of SiO2.
a) Which is limiting, which is excess?
b) How much excess reactant is present?
c) How many moles of each product?
Solution
a) Pick one: say we pick SiO2.
0.50 mol SiO2* 4 mol HF = 2.0 mol HF needed
1 mol SiO2
We have 3.5 mol of HF, thus it is in excess and
the SiO2 is limiting.
b) 3.5 mol – 2.0 mol = 1.5 mol excess HF.
c) SiO2 + 4HF  SiF4 + 2H2O
0.50 2.0       0.50 1.0
mol     mol           mol mol
Example 4
36.0 g of H2O is added to 167 g of Fe. The
reaction is:
3Fe + 4 H2O  Fe3O4 + 4H2.
a) Which is limiting, which is excess?
b) How much excess reactant is present?
c) What is the theoretical yield of each
product?
Part a
Pick one of the reactants:
Say H2O
Do mass  mass stoichiometry to determine
the amount of the Fe needed in the reaction.
36.0 g H2O * 1 mol H2O * 3 mol Fe * 55.845 g Fe =
18.0148 g 4 mol H2O      1 mol Fe
83.7 g Fe
What does this mean?
Need 83.7 g: have 167 g.
More than enough so Fe is in excess and H2O is
limiting.
Part b
How much excess Fe?
Have 167 g and 83.7 g reacts
Thus the amount of excess is:
167 g – 83.7 g = 83.3
Round to 83 g
Part c
Do mass  mass stoichiometry for your
products:
[36.0 g H2O * 1 mol H2O] * 4 mol H2 * 2.0158 g = 4.03 g
18.0148 g 4 mol H2O 1 mol H2

[ ] * 1 mol Fe3O4 * 231.531 g = 116 g Fe3O4
4 mol H2O     1 mol Fe3O4
Example 5
3.75 g of aluminum chloride reacts with 7.27
g of sodium sulfate.
a) Which is limiting, which is excess?
b) How much excess reactant is present?
c) What is the theoretical yield of each
product?
Example 5
a) AlCl3 is limiting

b) 1.28 g excess Na2SO4

c) 4.81 g Al2(SO4)3
4.93 g NaCl
Work for Example 5
3.75 g AlCl3 * 1 mol AlCl3 * 3 mol Na2SO4 * 142.041 g =
133.341 g 2 mol AlCl3       1 mol Na2SO4
5.99 g Na2SO4
Have 7.27 g Na2SO4 need 5.99 g. Have excess thus AlCl3 is
limiting.
7.27 g Na2SO4 * 1 mol Na2SO4 * 2 mol AlCl3 * 133.341 g =
142.041 g    3 mol Na2SO4 1 mol AlCl3
4.54 g AlCl3
Have 3.75 g AlCl3, need 4.54 g. Don’t have enough AlCl3 so it
is limiting and the Na2SO4 is excess.
Other work
b) 7.27 g – 5.99 g = 1.28 g Na2SO4

c)
3.75 g AlCl3 * 1 mol AlCl3 * 1 mol Al2(SO4)3 * 342.147 g =
133.341 g      2 mol AlCl3 1 mol Al2(SO4)3
4.81 g Al2(SO4)3
3.75 g AlCl3 * 1 mol AlCl3 * 6 mol NaCl * 58.443 g =
133.341 g 2 mol AlCl3 1 mol NaCl
4.93 g NaCl
Example 5 continued
Say you did this as an experiment and
collected 3.75 g of NaCl.
yield was supposed to be 4.93 g.
% yield = Actual * 100 = 3.75 *100 = 76.1%
Theoretical        4.93
Mol  Mol Stoichiometry
For the following reactions: write the
balanced equation and determine the
number of moles of each substance
involved in the reaction.
1) 1.50 mol of potassium reacts with
magnesium nitrate.
2) 1.25 mol of sodium phosphate reacts
with barium nitrate.
3) 4.50 moles of ethane (C2H6) burns.
2K + Mg(NO3)2  2KNO3 + Mg
1.50 0.750       1.50   0.750
mol   mol        mol    mol
Work
Mg(NO3)2 & Mg:
1.50 mol K * 1 mol Mg(NO3)2 = 0.750 mol
2 mol K
KNO3:
1.50 mol K * 2 mol KNO3 = 1.50 mol
2 mol K
2Na3PO4 + 3Ba(NO3)2  6NaNO3 + Ba3(PO4)2
1.25 mol 1.88 mol        3.75 mol 0.625 mol
Work:
1.25 mol Na3PO4 * 1 mol Ba3(PO4)2 = 0.625 mol
2 mol Na3PO4
0.625 mol Ba3(PO4)2 * 3 mol Ba(NO3)2 = 1.875 mol
1 mol Ba3(PO4)2
1.875 mol Ba(NO3)2 * 6 mol NaNO3 = 3.75 mol
3 mol Ba(NO3)2