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.
         Reading equations
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.
For each CaSO4 you start with you need 2
  Na. This ratio must be true.
         Reading equations
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
  What if we start with 2.50 moles of CaSO4?
  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
• If you start with mass you have to convert
  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?
      Answers for example 2
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:
But start with H2O since it is limiting.
[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
Answers:
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
If start with AlCl3:
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.
If start with Na2SO4:
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.
What is your % yield?
Based on your calculations your theoretical
 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.
                Answer #1
         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
                Answer #2
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
                Answer #3
2C2H6 + 7O2  4CO2 + 6H2O
4.50    15.8      9.00    13.5
mol     mol     mol      mol
Work:
4.50 mol C2H6 * 4 mol CO2 = 9.00 mol CO2
                2 mol C2H6
4.50 mol C2H6 * 6 mol H2O = 13.5 mol H2O
                2 mol C2H6
4.50 mol C2H6 * 7 mol O2 = 15.8 mol O2
                2 mol C2H6

								
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