Solutions Stoichiometry PowerPoint - Questions, Calculations

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Solutions Stoichiometry PowerPoint - Questions, Calculations Powered By Docstoc
					            Stoichiometry overview
•   Recall that in stoichiometry the mole ratio
    provides a necessary conversion factor:
      molar mass of x           molar mass of y
grams (x)  moles (x)  moles (y)  grams (y)

         mole ratio from balanced equation
•   We can do something similar with solutions:
         mol/L of x              mol/L of y
volume (x)  moles (x)  moles (y)  volume (y)

         mole ratio from balanced equation
•   Read pg. 351-353. Try Q 1-3a.
           Pg. 353, Question 1
Ammonium sulfate is manufactured by reacting
sulfuric acid with ammonia. What concentration
of sulfuric acid is needed to react with 24.4 mL
of a 2.20 mol/L ammonia solution if 50.0 mL of
sulfuric acid is used?
     H2SO4(aq) + 2NH3(aq)  (NH4)2SO4(aq)
 Calculate mol H2SO4, then mol/L = mol/0.0500 L
  # mol H2SO4=
0.0244 L NH3 x 2.20 mol NH3 x 1 mol H2SO4 =0.02684 mol
                  L NH3        2 mol NH3       H2SO4
 mol/L = 0.02684 mol H2SO4 / 0.0500 L = 0.537 mol/L
              Pg. 353, Question 2
  Calcium hydroxide is sometimes used in water
  treatment plants to clarify water for residential
  use. Calculate the volume of 0.0250 mol/L
  calcium hydroxide solution that can be
  completely reacted with 25.0 mL of 0.125 mol/L
  aluminum sulfate solution.
   Al2(SO4)3(aq) + 3Ca(OH)2(aq)  2Al(OH)3(s) + 3CaSO4(s)
   # L Ca(OH)2=
 0.0250 0.125 mol Al2(SO4)3 3 mol Ca(OH)2 L Ca(OH)2
          x                    x                x
L Al2(SO4)3 L Al2(SO4)3         1 mol Al2(SO4)3 0.0250 mol
                                                   Ca(OH)2
             = 0.375 L Ca(OH)2
             Pg. 353, Question 3
 A chemistry teacher wants 75.0 mL of 0.200
 mol/L iron(Ill) chloride solution to react
 completely with an excess of 0.250 mol/L
 sodium carbonate solution. What volume of
 sodium carbonate solution is needed?
 2FeCl3(aq) + 3Na2CO3(aq)  Fe2(CO3)3(s) + 6NaCl(aq)
  # L Na2CO3=
0.0750 L FeCl3 0.200 mol FeCl3 3 mol Na2CO3 L Na2CO3
               x               x             x
                   L FeCl3       2 mol FeCl3 0.250 mol
                                               Na2CO3
        = 0.0900 L Na2CO3 = 90.0 mL Na2CO3
                      Answers
    1. H2SO4(aq) + 2NH3(aq)  (NH4)2SO4(aq)
 Calculate mol H2SO4, then mol/L = mol/0.0500 L
   # mol H2SO4=
0.0244 L NH3 x 2.20 mol NH3 x 1 mol H2SO4 =0.02684 mol
                     L NH3       2 mol NH3        H2SO4
 mol/L = 0.02684 mol H2SO4 / 0.0500 L = 0.537 mol/L
 2. Al2(SO4)3(aq) + 3Ca(OH)2(aq)  2Al(OH)3(s) + 3CaSO4(s)
   # L Ca(OH)2=
 0.0250 0.125 mol Al2(SO4)3 3 mol Ca(OH)2 L Ca(OH)2
          x                    x               x
L Al2(SO4)3 L Al2(SO4)3        1 mol Al2(SO4)3 0.0250 mol
                                                   Ca(OH)2
            = 0.375 L Ca(OH)2
                     Answers
3. 2FeCl3(aq) + 3Na2CO3(aq)  Fe2(CO3)3(s) + 6NaCl(aq)
  # L Na2CO3=
0.0750 L FeCl3 0.200 mol FeCl3 3 mol Na2CO3 L Na2CO3
               x               x             x
                   L FeCl3       2 mol FeCl3 0.250 mol
                                               Na2CO3
        = 0.0900 L Na2CO3 = 90.0 mL Na2CO3
                Assignment
1. H2SO4 reacts with NaOH, producing water
   and sodium sulfate. What volume of 2.0 M
   H2SO4 will be required to react completely
   with 75 mL of 0.50 mol/L NaOH?
2. How many moles of Fe(OH)3 are produced
   when 85.0 L of iron(III) sulfate at a
   concentration of 0.600 mol/L reacts with
   excess NaOH?
3. What mass of precipitate will be produced
   from the reaction of 50.0 mL of 2.50 mol/L
   sodium hydroxide with an excess of zinc
   chloride solution.
                Assignment
4. a) What volume of 0.20 mol/L AgNO3 will be
   needed to react completely with 25.0 mL of
   0.50 mol/L potassium phosphate?
   b) What mass of precipitate is produced from
   the above reaction?
5. What mass of precipitate should result when
   0.550 L of 0.500 mol/L aluminum nitrate
   solution is mixed with 0.240 L of 1.50 mol/L
   sodium hydroxide solution?
                     Answers
1. H2SO4(aq) + 2NaOH(aq)  2H2O + Na2SO4(aq)
   # L H2SO4=
0.075 L NaOH x0.50 mol NaOH 1 mol H2SO4     L H2SO4
                           x            x
                  L NaOH     2 mol NaOH 2.0 mol H2SO4
                              = 0.009375 L = 9.4 mL
2. Fe2(SO4)3(aq) + 6NaOH(aq)  2Fe(OH)3(s) + 3Na2SO4(aq)
 # mol Fe(OH)3=
85 L Fe2(SO4)3 x 0.600 mol Fe2(SO4)3 x 2 mol Fe(OH)3
                     L Fe2(SO4)3      1 mol Fe2(SO4)3
                                      = 102 mol
3. 2NaOH(aq) + ZnCl2(aq)  Zn(OH)2(s) + 2NaCl(aq)
  # g Zn(OH)2=                               = 6.21 g
 0.0500 x 2.50 mol NaOH 1 mol Zn(OH)2 99.40 g Zn(OH)2
                        x            x
L NaOH        L NaOH      2 mol NaOH   1 mol Zn(OH)2
 4a. 3AgNO3(aq) + K3PO4(aq)  Ag3PO4(s) + 3KNO3(aq)
 # L AgNO3 =                     = 0.1875 L = 0.19 L
 0.025 0.50 mol K3PO4 3 mol AgNO3          L AgNO3
         x             x            x
L K3PO4     L K3PO4      1 mol K3PO4 0.20 mol AgNO3
  4b. 3AgNO3(aq) + K3PO4(aq)  Ag3PO4(s) + 3KNO3(aq)
  # g Ag3PO4=                                  = 5.2 g
 0.025 x0.50 mol K3PO4 1 mol Ag3PO4 418.58 g Ag3PO4
                       x            x
L K3PO4     L K3PO4     1 mol K3PO4 1 mol Ag3PO4
5. Al(NO3)3(aq) + 3NaOH(aq)  Al(OH)3(s) + 3NaNO3(aq)
# g Al(OH)3=
  0.550 0.500 mol Al(NO3)3 1 mol Al(OH)3 77.98 g Al(OH)3
        x                 x              x
L Al(NO3)3 L Al(NO3)3      1 mol Al(NO3)3 1 mol Al(OH)3
                                    =    21.4 g Al(OH)3
# g Al(OH)3=
  0.240 1.50 mol NaOH 1 mol Al(OH)3 77.98 g Al(OH)3
        x            x             x
L NaOH       L NaOH    3 mol NaOH 1 mol Al(OH)3
                               =    9.36 g Al(OH)3


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