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PartIV LQ

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					Structured Questions


|!|EL04001|!|
A student tried to extract iron metal from haematite by strong heating. However, he failed. He could not
explain the failure. He thought that silver, also a kind of metal, could be obtained by heating its ore, so
iron could also be extracted by the same method.       [10]
a.      Explain the extraction failure and the misunderstanding of the student.
b.      Suggest a method to extract iron from haematite.
c.      Write a balanced chemical equation for the reaction involved in your method stated in (b).
d.      The student said, “The bonds in iron and haematite are not the same, but they are both formed from
        the electrostatic force between oppositely charged particles”. Explain what he said.
e.      The student at last extracted iron metal from haematite successfully. He found that the colour of
        iron metal is very different from that of haematite. Why?
##
a.      Metals that situated above silver in the reactivity series cannot be extracted by heating its ores. Iron
        is more reactive than silver, therefore it cannot be extracted by strong heating alone. The extraction
        method stated above failed.    (1)
b.      Heat haematite with carbon or carbon monoxide at about 1000°C / high temperature.               (1)
c.      2Fe2O3(s) + 3C(s)  4Fe(s) + 3CO2(g) /
        Fe2O3(s) + 3CO(g)  2Fe(s) + 3CO2(g) (1)
d.      The bond in iron metal is metallic bond (1). In metals, the outermost electrons of atoms form a sea
        of delocalized electrons (1), leaving the atoms as cations (1). The electrostatic attraction force
        between cations and the electron sea holds the particles together. (1). The bond in haematite,
        iron(III) oxide, is ionic bond (1). The electrostatic forces between the positively charged iron(III)
        ions and the negatively charged oxide ions help to hold the particles together.     (1)
e.      The colour of haematite is due to the presence of iron(III) ions, not iron.   (1)
##


|!|EL04002|!|
Answer the following questions with the information given in the table.        [9]

                                    Aluminium                       Iron                      Gold

 % abundance by mass
                                        8.1                         5.0                     0.0000004
      in the Earth’s crust

     World production per
                                       15000                     716000                           1.4
 year (thousand tonnes)

 Approximate price per
                                       10500                        900                     82000000
         tonne (HK$)
a.      Why the price of aluminium is higher than that of iron, although aluminium is more abundant than
      iron?
b.    Account for the high price of gold.
c.    Suggest a reason why, aluminium, instead of iron, is usually used in making window frames.
d.    A man would like to use iron to make window frames as iron is much cheaper. Suggest a method to
      prevent the frames from rusting.
e.    Recycling helps saving metal reserve in the Earth’s crust, but people sometimes would like to
      extract metals rather than recycling. Why?
##
a.    Aluminium is extracted by electrolysis of its molten ore, while iron is extracted by using a blast
      furnace. (1) The cost of operating electrolysis is much higher than that of using a blast furnace. (1)
b.    Gold is a rare metal (1), and the demand for it is high as it is an unreactive metal and can be used
      for making ornaments, jewellries and protective coatings. (1). With low supply but high demand,
      the price for gold is high.
c.    Aluminium is more corrosion-resistant. (1) Aluminium reacts with oxygen in air to form a
      nonporous oxide layer (1), protecting aluminium from corrosion; iron alone is subjected to the
      attack of oxygen and water. It forms porous rust and therefore it goes further rusting easily.   (1)
d.    Painting (1). It can keep the iron away from oxygen and moisture.
e.    The cost of recycling is sometimes higher than the extraction cost.    (1)
##


|!|EL04003|!|
A student added excess zinc powder into a beaker of copper(II) sulphate solution. He found that there was
a colour change in the solution.    [10]
a.    Describe and explain the colour change in the solution.
b.    Write a balanced equation for the reaction.
c.    What would happen to the observation of colour change if a zinc bar was used instead of zinc
      powder in the experiment? Explain your answer.
d.    What would happen if silver powder is added to the product solution of the above experiment?
      Explain.
##
a.    The colour of the solution changed from blue to colourless gradually. (1) As zinc is more reactive
      than copper, a displacement reaction took place (1): blue copper(II) ions were reduced to copper
      metal gradually (1); while zinc gradually dissolved to form colourless zinc ions.    (1)
b.    Zn(s) + CuSO4(aq)  ZnSO4(aq) + Cu(s) (1)
c.    The colour change would not be observed (1). As copper formed in the displacement reaction would
      deposit on the surface of the zinc bar (1), the reaction would then stop.    (1)
      OR      The blue colour would disappear very slowly (1)
              As the surface area for the reaction to take place would be greatly reduced if zinc powder
              were replaced with zinc bar (1), the reaction would be very slow (1), leading to slow colour
              change.
d.     There is no reaction (1) because silver is less reactive than zinc, it cannot displace zinc.   (1)
##


|!|EL04004|!|
The table summarizes some information about three metals, X, Y and Z. [12]

                                                      X                    Y                      Z

                                               A white solid     A white solid formed        A black solid
          Reaction with oxygen
                                                    formed                                     formed

        Reaction with cold water                  Gas evolved         Gas evolved            No reaction

 Reaction with dilute hydrochloric acid           Gas evolved          Explosion             Gas evolved

a.     Arrange the three metals in descending order of reactivity.
b.     How Y is stored in laboratory?
c.     Y is a Group I metal. Write a balanced equation for the reaction of metal Y and water.
d.     State the observation when metal Y reacts with water.
e.     What would metal X be? Write a balanced equation for the reaction between metal X and dilute
       hydrochloric acid, using the metal name you stated.
f.     A piece of metal X is burnt in Bunsen flame. What is the colour of the flame?
g.     Suggest what metal Z would be. By what method is metal Z extracted from its ore? Write an
       equation for the reaction involved in the extraction.
##
a.     Y > X > Z (1)
b.     Y is stored under paraffin oil in laboratory. (1)
c.     2Y(s) + 2H2O(l) 2YOH(aq) + H2(g) (1)
d.     The metal moves quickly around the water surface (1) and burns with coloured flames. (1) It melts
       and dissolves in water gradually.    (1)
e.     Calcium; (1) Ca(s) + 2HCl(aq)  CaCl2(aq) + H2(g) (1)
f.     Brick red.   (1)
g.     Iron. (1) Iron is extracted by carbon reduction of its ore.   (1)
       2Fe2O3(s) + 3C(s)  4Fe(s) + 3CO2(g) (1)
##


|!|EL04005|!|
Copper can be extracted from copper pyrites, an ore containing copper(II) sulphide, by carbon reduction.
[11]
a.     Suggest a solid reducing agent used in the extraction.
b.     Write balanced equation(s) to illustrate the reaction(s) in the extraction process.
c.     Explain why lead extraction requires a higher temperature than copper extraction does.
d.     Suggest a method to extract copper other than by carbon reduction.
Extension
e.    How many grams of reducing agent suggested in part (a) is required to extract all copper from
      8.604 g of copper(II) sulphide? (Relative atomic masses: C= 12.0; O= 16.0; S= 32.1; Cu= 63.5)
##
a.    Carbon / coke (1)
b.    2CuS(s) + 3O2(g)  2CuO(s) + 2SO2(g) (1)
      2CuO(s) + C(s)  2Cu(s) + CO2(g) (1)
c.   Lead is more reactive than copper, (1) thus lead compounds are more stable than copper compounds.
      (1) As a result, more energy, and hence higher temperature, is needed to convert lead compounds
      into lead. (1)
d.    Add zinc powder into copper(II) sulphate solution. (1)
      Or    any correct displacement reactions.
Extension
e.    Consider the following equations:
      2CuS(s) + 3O2(g)  2CuO(s) + 2SO2(g) …………..(I)
      2CuO(s) + C(s)  2Cu(s) + CO2(g) ….……………(II)
      Number of moles of CuS(s) present = 8.604/(63.5+32.1) = 0.09 mol       (1)
      From equation (I), mole ratio of CuS to CuO = 1:1
       number of moles of CuO formed = 0.09 mol (1)
      From equation (II), mole ratio of CuO to C = 2:1
       number of moles of C needed = 0.045 mol (1)
      Mass of C needed = 0.045 x 12.0 = 0.54 g (1)


|!|EL04006|!|
In an experiment, a sample of 24.08 g iron burns in oxygen to form an oxide.       [12]
a.    What is the formula of the oxide if the oxide contains 70% of iron by mass?
b.    Find the formula mass of the oxide.
Extension
c.    What is the mass of oxygen required for a complete combustion of the iron sample? Write down all
      steps in your calculation.
d.    The oxide is then reduced back to iron by carbon. Calculate the amount of carbon needed to obtain
      all iron in the product formed from the reaction stated in part (c).
      (The relative atomic masses: C= 12.0; O= 16.0; Fe= 56.0)
##
a.    Let there be 100 g of the iron oxide.
      Mole ratio of Fe : O = 70/56 : (100-70)/16     (1)
                            = 1.25 : 1.875
                            =2:3     (1)
      The formula of the oxide is Fe2O3.      (1)
b.    The formula mass of Fe2O3
         = 2  56 + 3  16
         = 160      (1)
Extension
c.       The equation for the reaction is
                                  4Fe(s)              +     3O2(g)         2Fe2O3(s)   (1)
         Mole ratio                     4             :       3
         No. of moles             24.08/ 56                   0.43  3/4
                                  = 0.43 (1)                  = 0.3225     (1)
         The mass of oxygen required for complete combustion
         = 0.3225  (16  2)
         = 10.32 g        (1)
d.       Refer to the equation in part (c), the amount of Fe2O3 produced is 0.43/ 2 = 0.215 mol.
         (1)
         The equation for the reduction is
                                2Fe2O3(s) + 3C(s)  4Fe(s) + 3CO2(g)             (1)
         Mole ratio               2         :     3
         No. of mole            0.215           0.215  3/2
                                                = 0.3225    (1)
         The mass of carbon required for complete reduction
         = 0.3225  12.0
         = 3.87 g     (1)
##


|!|EL04007|!|
A student lost the labels of three bottles, and each of which contains one kind of metal. In order to find
out what the metals (they are temporarily called P, Q and R respectively) are, he placed these metals in
different solutions to see if displacement reactions occur. The following table summarizes the results of
the experiments. In the table, a (+) sign indicates that a displacement reaction occurred; while a () sign
indicates that no reaction occurred. [15]

 Metal         MgCl2(aq)              FeCl3(aq)            CuCl2(aq)

     P               +                      +                     +

     Q                                                          +

     R                                     +                     +

a.       What metals would P, Q and R possibly be? Explain your answer.
b.       The experiments conducted do not provide sufficient information to identify each metal. Suggest
         experiments to identify them based on your answer in part (a).
c.       In part (a), positions of metals in the reactivity series can be predicted approximately by those
         displacement reactions. Can the method of placing those metals in dilute hydrochloric acid be used
          in laboratory for the same purpose?
##
a.        P is more reactive than magnesium. (1) It would be potassium, sodium or calcium; (1) Q is more
          reactive than copper but less reactive than iron. (1) It would be lead; (1) R is more reactive than iron
          but less reactive than magnesium. (1) It would be aluminium or zinc. (1)
b.        P may be potassium, sodium or calcium. Place a piece of P in cold water. (1) Potassium and sodium
          will move around on the water surface quickly. Potassium will give a lilac flame, while sodium will
          give a golden yellow flame; (2) calcium will sink in water and no flame will be produced.         (1)
          Or      Carry out a flame test to test P. (1) Potassium will give a lilac flame; sodium will give a
          golden yellow flame; calcium will give a brick red flame.(3) \
          R may be aluminium or zinc. Pass steam through a sample of R. (1) If gas is evolved, it may be zinc
          as zinc reacts with steam to produce hydrogen gas; if no gas is evolved, it may be aluminium as
          aluminium has a protective oxide layer on its surface to prevent it from reacting with steam.         (1)
c.        No. It is very dangerous (1) to put unknown metals in dilute hydrochloric acid, as some metals,
          such as potassium and sodium, can cause explosion when reacting with acid.          (1)
##


|!|EL04008|!|
Barium (Ba), with atomic number 56, is in the same group with calcium and magnesium in the Periodic
Table.      [9]
a.        Which metal, calcium or barium, is more reactive? Explain your answer.
b.        Write a balanced equation for the reaction between barium and dilute hydrochloric acid.
c.        By what method is barium extracted from its ore?
d.        Can barium be used to prevent iron from rusting? Why?
##
a.        Barium is more reactive.        (1)
          Barium has a lower position than calcium in the Periodic Table. (1) The lower a metal element
          situates in the group, the more reactive it is.    (1)
b.        Ba(s) + 2HCl(aq)  BaCl2(aq) + H2(g)         (1)
c.        Barium is extracted by electrolysis of its molten ore.     (1)
d.        Yes. (1) Barium should be more reactive than iron because it is more reactive than calcium, which
          is at a higher position than iron in the metal reactivity series. (1) Barium can prevent iron rusting by
          sacrificial protection.   (1)
##


|!|EL04009|!|
XXX is a hot city located near the coast. Its air quality is poor due to overpopulation and intense
industrial development. Her citizens found that articles made of iron rust much faster than those in other
cities.     [16]
a.        Explain this phenomenon.
b.    Suggest three ways to slow down iron rusting in the city.
c.    A man bought a new aluminium cooking pan. A few days later, he found that there was a thin white
      layer covering the pan surface. He scratched the layer off but it appeared again after a few days. He
      then scratched off the layer once it formed. After a period of time, the pan broke. Explain the
      phenomenon with appropriate chemical equation(s).
d.    Anodization is a well-grown industry in the city.
      i)     State the main function of anodization.
      ii)    Describe how anodization can be done in laboratory.
##
a.    Air near the coast contains many electrolytes, such as sodium chloride, which can accelerate
      corrosion by providing a better conducting medium (for reaction of iron with water and oxygen).
      (1)
      High temperature causes high reaction rate, making rusting goes faster.                (1)
      Overpopulation and intense industrial development produces many pollutants in the air. Common
      air pollutants are sulphur dioxide, nitrogen oxides and carbon monoxide. These air pollutants
      dissolve in rainwater to form acid, which provides hydrogen ions, and then corrodes iron.      (1)
b.    Painting (1)
      Electroplating (1)
      Sacrificial protection (1) or any other correct methods
c.    When the new aluminium cooking pan was in use, the oxygen in air reacted with aluminium to
      form a white aluminium oxide layer covering the pan surface.          (1)
      4Al(s) + 3O2(g)  2Al2O3(s) (1)
      After the layer was scratched off, the clean aluminium continued to react with oxygen and form
      oxide layer again. (1) The repeated scratching caused more aluminium reacting with oxygen and
      finally all aluminium had reacted and the pan broke.       (1)
d.    (i)    Anodization is to thicken the oxide layer on the surface of aluminium (1), and hence to
             enhance its resistance to corrosion.    (1)
      (ii)   In a setup of electrolysis, the aluminium object to be anodized is placed at the positive
             electrode (1), and the negative electrode (1) is an aluminium sheet, using dilute sulphuric acid
             as the electrolyte. (1) The oxygen evolved at the positive electrode reacts with aluminium to
             increase the thickness of the oxide layer.    (1)
##


|!|EL04010|!|
Five metals (P, Q, R, S, and T) were tested in the following experiments.             [14]
(i)   Small pieces of the five metals were added to dilute hydrochloric acid and the volumes of gas
      collected in the first two minutes were recorded.

                       Metal           P        Q          R           S          T

                     Volume/cm3       53        90         0           32     75
(ii)   Steam was allowed to pass through small pieces of the five metals and the volumes of gas collected
       in the first two minutes were recorded.

                       Metal            P        Q         R         S         T

                    Volume/cm3          49       103       0         0         0

a.     Arrange the five metals in descending order of reactivity. Explain your answer.
b.     Suggest and explain what T is.
c.     Q is a Group II element. Write the equations for the reactions of Q with steam and with dilute
       hydrochloric acid.
d.     Describe a test for the gas formed in experiment (i).
##
a.     Q > T > P > S > R (2)
       Q is the most reactive as the largest volume of gas was evolved in the two reactions. (1) R is the
       least reactive as no reaction occurred in both experiments. (1) T is more reactive than P because T
       gave more gases in experiment (i). (1) P is more reactive than S because P gave more gases in
       experiment (i) and S had no reaction in experiment (ii).     (1)
b.     T is aluminium. (1) P reacts with both dilute hydrochloric acid and steam, so P is above lead in the
       reactivity series. (1) From part (a), T is more reactive than P, so T is above lead in the series. (1)
       That T did not react with steam was not due to its low position in the series but the nonporous
       protective oxide layer of aluminium prevents the attack from steam. Among the reactive metals,
       only aluminum has a protective oxide layer on its surface.        (1)
c.     Q(s) + H2O(g)  QO(s) + H2(g) (1)
       Q(s) + 2HCl(aq)  QCl2(aq) + H2(g) (1)
d.     Place a burning splint near the mouth of the test tube containing the gas. (1) The gas is hydrogen if
       the splint burns with a “pop” sound.     (1)
##


|!|EL04011|!|
Silver is formed if zinc powder or magnesium powder is added to silver nitrate solution.         [9]
a.     Adding which powder, zinc or magnesium, would the process of silver formation be faster?
b.     Explain why magnesium powder helps obtaining silver faster than magnesium ribbon does.
c.     How many grams of magnesium powder should be added into silver nitrate solution to obtain 5 g of
       silver?
d.     If 0.8 g of zinc reacts with excess silver nitrate solution to form silver. What mass of silver can be
       obtained?
(Relative atomic masses: N=14.0; O=16.0; Mg=24.3; Zn=65.4; Ag = 107.9;)
##
a.     Magnesium powder (1). As magnesium is more reactive than zinc (1), magnesium undergoes
       displacement reaction faster.
b.    Magnesium powder provides greater surface area for the reaction to occur.            (1)
c.    The ionic equation for the reaction is
                             Mg(s) + 2Ag+(aq)  Mg2+(aq) + 2Ag(s)         (1)
      Mole ratio                 1             :                2
      No. of moles           0.0463/2                         5/107.9
                             = 0.0232                         = 0.0463        (1)
      Mass of magnesium required to obtain 5 g of silver
      = 0.0232  24.3
      = 0.5634g     (1)
d.    The equation for the reaction is
                             Zn(s) + 2AgNO3(aq)  Zn(NO3)2(aq) + 2Ag(s)             (1)
      Mole ratio         :      1                  :                      2
      No. of moles           0.8/65.4                                   0.0122  2
                             = 0.0122                                   = 0.0245     (1)
      The mass of silver obtained
      = 0.0245  107.9
      = 2.64 g     (1)
##


|!|EL04012|!|
X, Y and Z are three different metal solids, having +2 charge when in compounds. The table below shows
the results of two experiments about the metals.       [15]

                 Experiment                Metal X             Metal Y               Metal Z

         Adding metal to dilute          Gas evolved          No reaction           Gas evolved
            hydrochloric acid

      Adding metal to cold water         No reaction          No reaction           Gas evolved

 a.   Arrange the three metals in ascending order of reactivity.
 b.   Which metal forms the most stable oxide? Explain your answer.
 c.   Suggest and explain what metal Y would be.
Extension
d.    When 0.42 g of molten oxide of metal Y undergoes electrolysis, it decomposes completely to give
      0.12 g of oxygen. Calculate the formula mass of the oxide and the relative atomic mass of Y.
##
a.    Y < X < Z (2)
b.    Z. (1) Z is the most reactive one and hence its oxide is most stable.         (1)
c.    Copper. (1) As Y does not react with both cold water and dilute hydrochloric acid, it must be lower
      than lead in the reactivity series. (1) The metals lower than lead are copper, mercury, silver and gold.
      (1) Mercury is liquid (1), the charge of silver in compound is usually +1 (1), and gold usually does
      not form any oxide at all (1). Copper is a solid and its charge in compound can be +2 (1). Therefore,
      it is reasonable to believe that Y is copper.
Extension
d.    No. of mole of oxygen atoms in YO = 0.12/16 = 0.0075                    (1)
      No. of mole of YO = no. of mole of oxygen atoms = 0.007                    (1)
      Formula mass of YO = 0.42/0.0075
                              = 56    (1)
     Relative atomic mass of Y = 56 –16 = 40                (1)
##


|!|EL04013|!|
A mixture of zinc sulphide and zinc oxide is heated. After strong heating for 30 minutes, 0.52 g of a
gaseous product A is formed. The residue is then heated strongly with excess carbon. 0.70 g of another
gaseous product B is formed after the reaction is complete. [13]
a.    Identify the two gaseous products in the reactions involved.
b.    Write balanced equations for the reactions as described above.
Extension
c.    Calculate the percentage by mass of zinc oxide in the mixture.
##
a.    A is sulphur dioxide. B is carbon dioxide.             (2)
b.    2ZnS(s) + 3O2(g)  2ZnO(s) + 2SO2(g)                  (1)
      2ZnO(s) + C(s)  2Zn(s) + CO2(g)                (1)
Extension
c.    The first reaction is
                                  2ZnS(s)         +         3O2(g)          2ZnO(s)     +        2SO2(g)
      Mole ratio                     1                        :                  1            :      1
      No. of moles                                                                        0.52/(32.1 + 16  2)       (1)
                                   0.0081                                    0.0081                 = 0.0081   (1)
      Mass                         0.0081  (65.4 + 32.1)                    0.0081  (65.4 + 16)
                                   = 0.79 g                                  = 0.66 g   (2)
The second reaction is
                          2ZnO(s) + C(s)  2Zn(s) + CO2(g)
      Mole ratio              2                   :                  1
      No. of moles        0.016  2                               0.7/(12 + 16 + 16)    (1)
                          = 0.032                                 = 0.016
      Mass                0.032  (65.4 + 16)
                          = 2.6048 g        (1)
The total mass of ZnO reacted is 2.6048 g. This includes the ZnO initially present in the mixture and the
product from ZnS. Therefore,
mass of ZnO initially present in the mixture = 2.6048 – 0.66 = 1.9448g (1)
The total mass of the mixture = mass of ZnS + mass of ZnO
                                     = 0.79 + 1.9448
                                     = 2.7348 g (1)
The percentage by mass of ZnO in the mixture
= (1.9448/2.7348)  100%
= 71.11 %    (1)
##


|!|EL04014|!|
Five iron nails are placed at different conditions. The setup is shown below. [11]
A.                                      B.
                         anhydrous
                         calcium chloride              oil layer

                                                       fresh distilled water
                         cotton wool
                                                       iron nail
                         iron nail
C.                                        D.                               E.
                   air
                                                       sodium chloride                      tap water
                   tap water                           solution
                                                                                           magnesium ribbon
                     iron nail                         iron nail                            iron nail



a.    What factors are essential for the rusting process of iron?
b.    What is the purpose of placing anhydrous calcium chloride in tube A?
c.    Why oil and fresh distilled water were used in tube B?
d.    If the air in tube C is replaced with car exhaust gas, would the iron nail in it rust faster? Explain
      your answer.
e.    Would the nail in tube B rust? Explain your answer.
f.    If the magnesium ribbon is replaced with magnesium powder, would the iron nail start rusting at an
      earlier time? Why?
##
a.    The presence of both oxygen (1) and water (1).
b.    It absorbs moisture in air (1), removing one essential factor, water, to prevent or slow down rusting.
      (1)
c.    Oil prevents oxygen dissolving into the water below; (1) fresh distilled water ensures no ions / no
      air present.       (1)
d.    Yes. (1) Car exhaust gas contains many acidic pollutants (1), such as sulphur dioxide, carbon
      dioxide and nitrogen oxides. When they dissolve in water, acid forms and the ions accelerate the
      rusting process.         (1)
e.    Yes (1), as there is some oxygen in the unboiled distilled water.    (1)
f.   Yes. (1) As magnesium powder provides greater surface area for its reaction with air, (1)
     magnesium would run out earlier and therefore the nail would start rusting earlier.   (1)
##

				
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