WAEC SSCE - Chemistry - 2001 (Essay)
Question 1 Report
(a)(i) Draw and label a simple cell for the electrolytic purification of copper.
(ii) Write can equation for the reaction at each electrode in (a)(i) above.
(iii) State with reason whether the Daniell cell is an electrolytic cell or an electrochemical cell.
(iv) What is the function of MnO\(_2\) in the Laclanche cell?
(b) Consider the following equation: MnO\(^-_4\) + 8H\(^+\) + xe\(^-\) \(\to\) Mn\(^{2+}\) + yH\(_2\)O. State the
(i) values of x and y;
(ii) oxidation state of Mn in MnO\(^-_4\).
(c)(i) List three factors that affect selective discharge of ions during electrolysis
(ii) State Faraday's second law of electrolysis.
(iii) A voltameter containing silver trioxonitrate(V) solution was connected in series to another voltameter containing copper (II) tetraoxosulphate(VI) solution. When a current ri 0.200 ampere was passed through the solutions, 0.780g of silver was deposited. Calculate the
I. mass of copper that would be deposited in the copper voltameter
II. quantity of electricity used and the time of current flow. [Cu = 63.5 ; Ag = 108; 1F = 96500C]
(a)(i) Electrolytic purification of copper
Impure copper is made the anode and pure copper is the cathode. Both are immersed in acidified copper(II) sulfate solution and connected to a direct-current supply. Pure copper is deposited on the cathode, while insoluble impurities settle as anode mud.
(ii) Electrode reactions
Anode (oxidation): \[\mathrm{Cu_{(s)} \rightarrow Cu^{2+}_{(aq)} + 2e^-}\]
Cathode (reduction): \[\mathrm{Cu^{2+}_{(aq)} + 2e^- \rightarrow Cu_{(s)}}\]
(iii) The Daniell cell is an electrochemical (galvanic) cell because it converts chemical energy from a spontaneous redox reaction into electrical energy.
(iv) In the Leclanché cell, \(\mathrm{MnO_2}\) acts as a depolariser. It removes hydrogen formed at the carbon electrode, thereby preventing polarisation and a fall in voltage.
(b)(i)
Balancing oxygen atoms gives \(y=4\), and balancing charge gives \(x=5\):
\[\mathrm{MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O}\]
Therefore, \(\boxed{x=5}\) and \(\boxed{y=4}\).
(ii) Let the oxidation state of Mn be \(z\):
\[z+4(-2)=-1\]
\[z=+7\]
Hence, the oxidation state of Mn in \(\mathrm{MnO_4^-}\) is \(\boxed{+7}\).
(c)(i) Factors affecting selective discharge of ions
- Position of the ions in the electrochemical series.
- Concentration of the ions in the electrolyte.
- Nature of the electrode.
(ii) Faraday's second law of electrolysis
When the same quantity of electricity passes through different electrolytes, the masses of substances liberated or deposited are proportional to their chemical equivalent masses.
(iii) Calculation
Since the voltameters are connected in series, the same quantity of electricity passes through both solutions.
For silver:
\[\mathrm{Ag^+ + e^- \rightarrow Ag}\]
\[\text{Moles of Ag deposited}=\frac{0.780}{108}=7.222\times10^{-3}\ \text{mol}\]
One mole of Ag requires one mole of electrons. Therefore:
\[\text{Moles of electrons}=7.222\times10^{-3}\ \text{mol}\]
I. Mass of copper deposited
\[\mathrm{Cu^{2+}+2e^-\rightarrow Cu}\]
\[\text{Moles of Cu}=\frac{7.222\times10^{-3}}{2}=3.611\times10^{-3}\ \text{mol}\]
\[\text{Mass of Cu}=3.611\times10^{-3}\times63.5=0.229\ \text{g}\]
\[\boxed{\text{Mass of copper deposited}=0.229\ \text{g}}\]
II. Quantity of electricity and time of flow
\[Q=nF=(7.222\times10^{-3})(96500)=696.9\ \text{C}\]
\[\boxed{Q\approx697\ \text{C}}\]
Since \(Q=It\):
\[t=\frac{Q}{I}=\frac{696.9}{0.200}=3484.5\ \text{s}\]
\[\boxed{t\approx3.48\times10^3\ \text{s}=58.1\ \text{min}}\]
Answer Details
(a)(i) Electrolytic purification of copper
Impure copper is made the anode and pure copper is the cathode. Both are immersed in acidified copper(II) sulfate solution and connected to a direct-current supply. Pure copper is deposited on the cathode, while insoluble impurities settle as anode mud.
(ii) Electrode reactions
Anode (oxidation): \[\mathrm{Cu_{(s)} \rightarrow Cu^{2+}_{(aq)} + 2e^-}\]
Cathode (reduction): \[\mathrm{Cu^{2+}_{(aq)} + 2e^- \rightarrow Cu_{(s)}}\]
(iii) The Daniell cell is an electrochemical (galvanic) cell because it converts chemical energy from a spontaneous redox reaction into electrical energy.
(iv) In the Leclanché cell, \(\mathrm{MnO_2}\) acts as a depolariser. It removes hydrogen formed at the carbon electrode, thereby preventing polarisation and a fall in voltage.
(b)(i)
Balancing oxygen atoms gives \(y=4\), and balancing charge gives \(x=5\):
\[\mathrm{MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O}\]
Therefore, \(\boxed{x=5}\) and \(\boxed{y=4}\).
(ii) Let the oxidation state of Mn be \(z\):
\[z+4(-2)=-1\]
\[z=+7\]
Hence, the oxidation state of Mn in \(\mathrm{MnO_4^-}\) is \(\boxed{+7}\).
(c)(i) Factors affecting selective discharge of ions
- Position of the ions in the electrochemical series.
- Concentration of the ions in the electrolyte.
- Nature of the electrode.
(ii) Faraday's second law of electrolysis
When the same quantity of electricity passes through different electrolytes, the masses of substances liberated or deposited are proportional to their chemical equivalent masses.
(iii) Calculation
Since the voltameters are connected in series, the same quantity of electricity passes through both solutions.
For silver:
\[\mathrm{Ag^+ + e^- \rightarrow Ag}\]
\[\text{Moles of Ag deposited}=\frac{0.780}{108}=7.222\times10^{-3}\ \text{mol}\]
One mole of Ag requires one mole of electrons. Therefore:
\[\text{Moles of electrons}=7.222\times10^{-3}\ \text{mol}\]
I. Mass of copper deposited
\[\mathrm{Cu^{2+}+2e^-\rightarrow Cu}\]
\[\text{Moles of Cu}=\frac{7.222\times10^{-3}}{2}=3.611\times10^{-3}\ \text{mol}\]
\[\text{Mass of Cu}=3.611\times10^{-3}\times63.5=0.229\ \text{g}\]
\[\boxed{\text{Mass of copper deposited}=0.229\ \text{g}}\]
II. Quantity of electricity and time of flow
\[Q=nF=(7.222\times10^{-3})(96500)=696.9\ \text{C}\]
\[\boxed{Q\approx697\ \text{C}}\]
Since \(Q=It\):
\[t=\frac{Q}{I}=\frac{696.9}{0.200}=3484.5\ \text{s}\]
\[\boxed{t\approx3.48\times10^3\ \text{s}=58.1\ \text{min}}\]
Question 2 Report
(a) State three characteristic properties of
(i) electrovalent compounds;
(ii) alpha particles
(iii) catalysts
(b)(i) Write the electronic configuration of silicon (atomic number 14) and state the group to which it belongs in the Periodic Table.
(ii) State the type of chemical bonding between silicon and oxygen in SiO\(_2\)
(iii) A chip used in a microcomputer contains 5.72 x 10\(^{-3}\)g of silicon, calculate the number of silicon atoms in the chip.
[Si = 28; Avogadro constant = 6.02 x 10\(^{23}\) mol\(^{-1}\)]
(c) An element X belongs to the same group as sodium but is more reactive.
(i) Suggest with reason whether X would be a reducing or oxidizing agent.
(ii) What would be a suitable method of storing X in the laboratory?
(iii) Describe briefly what would be observed if a small piece of X were dropped into a trough of cold water which had been coloured with red litmus.
(iv) Write an equation to show how the oxide of X would react with dilute HCI.
(v) Suggest the likely colour of the salts of X
(a) Characteristic properties
(i) Electrovalent (ionic) compounds:
- High melting and boiling points.
- Conduct electricity when molten or dissolved in water.
- Usually soluble in water but insoluble in organic solvents.
(ii) Alpha particles:
- Positively charged (helium nuclei, 2 protons and 2 neutrons).
- Deflected by electric and magnetic fields (toward the negative plate).
- Low penetrating power (stopped by a sheet of paper) but strongly ionising.
(iii) Catalysts:
- Alter the rate of a reaction without being used up.
- Needed only in small amounts and are chemically unchanged at the end.
- Are specific and can be poisoned by impurities.
(b) Silicon
(i) Electronic configuration of Si (Z = 14): \(1s^2\,2s^2\,2p^6\,3s^2\,3p^2\), i.e. 2, 8, 4. It belongs to Group IV (Group 14).
(ii) Bonding between silicon and oxygen in SiO2: covalent (a giant covalent structure).
(iii) Number of silicon atoms:
Moles of Si \( = \dfrac{5.72 \times 10^{-3}}{28} = 2.043 \times 10^{-4}\ \text{mol} \)
Number of atoms \( = 2.043 \times 10^{-4} \times 6.02 \times 10^{23} = \mathbf{1.23 \times 10^{20}\ atoms} \)
(c) Element X (same group as Na but more reactive)
(i) X is a reducing agent. Being more electropositive than sodium, it loses its outer electron even more readily, so it reduces other species while being oxidised itself.
(ii) Store X under paraffin oil (kerosene) to keep it away from air and moisture.
(iii) If dropped into cold water coloured with red litmus, X would float and dart about on the surface, melt into a shining ball, hiss and give off a gas (hydrogen) which may ignite; the red litmus turns blue as an alkaline hydroxide is formed.
(iv) \[ X_2O + 2HCl \to 2XCl + H_2O \]
(v) The salts of X are likely to be white (colourless), as for other Group I salts.
Answer Details
(a) Characteristic properties
(i) Electrovalent (ionic) compounds:
- High melting and boiling points.
- Conduct electricity when molten or dissolved in water.
- Usually soluble in water but insoluble in organic solvents.
(ii) Alpha particles:
- Positively charged (helium nuclei, 2 protons and 2 neutrons).
- Deflected by electric and magnetic fields (toward the negative plate).
- Low penetrating power (stopped by a sheet of paper) but strongly ionising.
(iii) Catalysts:
- Alter the rate of a reaction without being used up.
- Needed only in small amounts and are chemically unchanged at the end.
- Are specific and can be poisoned by impurities.
(b) Silicon
(i) Electronic configuration of Si (Z = 14): \(1s^2\,2s^2\,2p^6\,3s^2\,3p^2\), i.e. 2, 8, 4. It belongs to Group IV (Group 14).
(ii) Bonding between silicon and oxygen in SiO2: covalent (a giant covalent structure).
(iii) Number of silicon atoms:
Moles of Si \( = \dfrac{5.72 \times 10^{-3}}{28} = 2.043 \times 10^{-4}\ \text{mol} \)
Number of atoms \( = 2.043 \times 10^{-4} \times 6.02 \times 10^{23} = \mathbf{1.23 \times 10^{20}\ atoms} \)
(c) Element X (same group as Na but more reactive)
(i) X is a reducing agent. Being more electropositive than sodium, it loses its outer electron even more readily, so it reduces other species while being oxidised itself.
(ii) Store X under paraffin oil (kerosene) to keep it away from air and moisture.
(iii) If dropped into cold water coloured with red litmus, X would float and dart about on the surface, melt into a shining ball, hiss and give off a gas (hydrogen) which may ignite; the red litmus turns blue as an alkaline hydroxide is formed.
(iv) \[ X_2O + 2HCl \to 2XCl + H_2O \]
(v) The salts of X are likely to be white (colourless), as for other Group I salts.
Question 3 Report
(a)(i) Explain what is meant by acid anhydride and give one example
(ii) State three chemical properties of hydrochloric acid.
(b) Explain each of the following observations:
(i) Tetraoxosulphate (VI) acid can form two types of salts unlike trioxonitrate (V) acid.
(ii) Copper and iron react with concentrated H\(_2\)SO\(_4\) but only one of them reacts with the dilute acid.
(iii) On adding dilute H\(_2\)SO\(_4\) separately to zinc dust and zinc granules of the same mass, the dust produced more vigorous effervescence.
(c)(i) Define activation energy.
(ii) Sketch and label an energy profile diagram for the following reaction: A + B --> C + D; AH = xkJmol\(^{-1}\)
(iii) Explain why the heat of reaction of the mineral acids with sodium hydroxide is constant in value.
(d) Consider the reaction represented by the following equation:
Q\(_{(s)}\) \(\rightleftharpoons\) Q\(_{(l)}\) \(\Delta\) = xkJmol\(^{-1}\)
(i) State with reason which of Q\(_{(s)}\) and O\(_{(J)}\) has the higher entropy.
(ii) What will be the effect of decrease in temperature on the system at equilibrium?
(a)(i) An acid anhydride is an oxide which reacts with water to form an acid. For example:
\[\mathrm{SO_3(g)+H_2O(l)\rightarrow H_2SO_4(aq)}\]
Thus, \(\mathrm{SO_3}\) is an acid anhydride.
(ii) Three chemical properties of hydrochloric acid are:
- It reacts with metals above hydrogen in the electrochemical series to form a chloride salt and hydrogen gas:
\[\mathrm{Zn(s)+2HCl(aq)\rightarrow ZnCl_2(aq)+H_2(g)}\] - It neutralises bases or alkalis to form a chloride salt and water:
\[\mathrm{HCl(aq)+NaOH(aq)\rightarrow NaCl(aq)+H_2O(l)}\] - It reacts with carbonates or hydrogencarbonates to produce carbon(IV) oxide, water and a chloride salt:
\[\mathrm{Na_2CO_3(aq)+2HCl(aq)\rightarrow 2NaCl(aq)+H_2O(l)+CO_2(g)}\]
(b)(i) \(\mathrm{H_2SO_4}\) is dibasic because it has two replaceable hydrogen ions. It can therefore form an acid salt, for example \(\mathrm{NaHSO_4}\), and a normal salt, for example \(\mathrm{Na_2SO_4}\). \(\mathrm{HNO_3}\) is monobasic, having only one replaceable hydrogen ion, so it forms only normal salts such as \(\mathrm{NaNO_3}\).
(ii) Concentrated \(\mathrm{H_2SO_4}\) is an oxidising acid. It reacts with both iron and copper, including copper which is below hydrogen in the electrochemical series. For example:
\[\mathrm{Cu(s)+2H_2SO_4(conc)\rightarrow CuSO_4(aq)+SO_2(g)+2H_2O(l)}\]
Dilute \(\mathrm{H_2SO_4}\) behaves as an ordinary acid. Iron, being above hydrogen in the electrochemical series, displaces hydrogen from it, whereas copper, being below hydrogen, does not:
\[\mathrm{Fe(s)+H_2SO_4(aq)\rightarrow FeSO_4(aq)+H_2(g)}\]
(iii) Zinc dust has a much larger surface area than zinc granules of the same mass. Hence, more zinc particles are in contact with the acid at a time, giving a faster reaction and more vigorous effervescence.
(c)(i) Activation energy is the minimum energy which reacting particles must possess for collisions between them to result in a chemical reaction.
(ii) Energy profile for \(\mathrm{A+B\rightarrow C+D}\), where \(\Delta H=+x\ \mathrm{kJ\,mol^{-1}}\):
(iii) Mineral acids such as hydrochloric acid, nitric acid and tetraoxosulphate(VI) acid are strong acids and are fully ionised in dilute aqueous solution. Their reaction with sodium hydroxide has the same net ionic equation:
\[\mathrm{H^+(aq)+OH^-(aq)\rightarrow H_2O(l)}\]
Since the same process, the formation of one mole of water, occurs in each case, the heat of neutralisation is approximately constant, about \(-57\ \mathrm{kJ\,mol^{-1}}\).
(d)(i) \(\mathrm{Q_{(l)}}\) has the higher entropy. In the liquid state, the particles have greater freedom of movement and a more random arrangement than in the solid state.
(ii) A decrease in temperature favours the reverse reaction, that is, the conversion of \(\mathrm{Q_{(l)}}\) to \(\mathrm{Q_{(s)}}\). The forward reaction is melting and is endothermic, so lowering the temperature favours the exothermic reverse change.
Answer Details
(a)(i) An acid anhydride is an oxide which reacts with water to form an acid. For example:
\[\mathrm{SO_3(g)+H_2O(l)\rightarrow H_2SO_4(aq)}\]
Thus, \(\mathrm{SO_3}\) is an acid anhydride.
(ii) Three chemical properties of hydrochloric acid are:
- It reacts with metals above hydrogen in the electrochemical series to form a chloride salt and hydrogen gas:
\[\mathrm{Zn(s)+2HCl(aq)\rightarrow ZnCl_2(aq)+H_2(g)}\] - It neutralises bases or alkalis to form a chloride salt and water:
\[\mathrm{HCl(aq)+NaOH(aq)\rightarrow NaCl(aq)+H_2O(l)}\] - It reacts with carbonates or hydrogencarbonates to produce carbon(IV) oxide, water and a chloride salt:
\[\mathrm{Na_2CO_3(aq)+2HCl(aq)\rightarrow 2NaCl(aq)+H_2O(l)+CO_2(g)}\]
(b)(i) \(\mathrm{H_2SO_4}\) is dibasic because it has two replaceable hydrogen ions. It can therefore form an acid salt, for example \(\mathrm{NaHSO_4}\), and a normal salt, for example \(\mathrm{Na_2SO_4}\). \(\mathrm{HNO_3}\) is monobasic, having only one replaceable hydrogen ion, so it forms only normal salts such as \(\mathrm{NaNO_3}\).
(ii) Concentrated \(\mathrm{H_2SO_4}\) is an oxidising acid. It reacts with both iron and copper, including copper which is below hydrogen in the electrochemical series. For example:
\[\mathrm{Cu(s)+2H_2SO_4(conc)\rightarrow CuSO_4(aq)+SO_2(g)+2H_2O(l)}\]
Dilute \(\mathrm{H_2SO_4}\) behaves as an ordinary acid. Iron, being above hydrogen in the electrochemical series, displaces hydrogen from it, whereas copper, being below hydrogen, does not:
\[\mathrm{Fe(s)+H_2SO_4(aq)\rightarrow FeSO_4(aq)+H_2(g)}\]
(iii) Zinc dust has a much larger surface area than zinc granules of the same mass. Hence, more zinc particles are in contact with the acid at a time, giving a faster reaction and more vigorous effervescence.
(c)(i) Activation energy is the minimum energy which reacting particles must possess for collisions between them to result in a chemical reaction.
(ii) Energy profile for \(\mathrm{A+B\rightarrow C+D}\), where \(\Delta H=+x\ \mathrm{kJ\,mol^{-1}}\):
(iii) Mineral acids such as hydrochloric acid, nitric acid and tetraoxosulphate(VI) acid are strong acids and are fully ionised in dilute aqueous solution. Their reaction with sodium hydroxide has the same net ionic equation:
\[\mathrm{H^+(aq)+OH^-(aq)\rightarrow H_2O(l)}\]
Since the same process, the formation of one mole of water, occurs in each case, the heat of neutralisation is approximately constant, about \(-57\ \mathrm{kJ\,mol^{-1}}\).
(d)(i) \(\mathrm{Q_{(l)}}\) has the higher entropy. In the liquid state, the particles have greater freedom of movement and a more random arrangement than in the solid state.
(ii) A decrease in temperature favours the reverse reaction, that is, the conversion of \(\mathrm{Q_{(l)}}\) to \(\mathrm{Q_{(s)}}\). The forward reaction is melting and is endothermic, so lowering the temperature favours the exothermic reverse change.
Question 4 Report
(a)(i) State three characteristics of a homologous series.
(ii) Give the name and structural formula of the second member of the alkyne series.
(iii) Write an equation to represent the combustion of ethane in excess oxygen.
(b) Name the type of reaction involved in the conversion of ethanol to
(i) ethene;
(ii) ethylethanoate;
(iii) chloroethane;
(iv) ethoxide
(v) ethanoic acid
(c) Consider the following compound.
(i) Write its IUPAC name.
(ii) Give its molecular formula and empirical formula
(iii) List the products of its H H 0 reaction with saturated Na\(_2\)CO\(_3\) solution.
(iv) State with reason whether its boiling point will be higher or lower than that of the corresponding alkane.
(d) A vegetable oil X was treated with activated charcoal and then with a gas Y in the presence of a catalyst in order to manufacture
(i) Identify Y.
(ii) State the function of the activated charcoal.
(iii) What is the catalyst used?
(iv) If a sample of X is heated with concentrated sodium hydroxide solution, list the products that will be obtained.
Answer Details
None
Question 5 Report
(a) List two substances that can be used in the laboratory to
(i) dry hydrogen;
(ii) remove carbon (IV) oxide from a sample of air;
(iii) convert hot copper (II) oxide to copper;
(iv) prepare zinc chloride by the action of dilute HCI.
(b)(i) Name two alloys which contain lead.
(ii) State and explain what is observed on bubbling \( \mathrm{H_2S} \) into a solution of \( \mathrm{Pb(NO)_2} \).
(iii) A metal M exists as a silvery white solid at temperatures above 18°C and as a grey solid below 18°C.
I. name the phenomenon exhibited by M.
II. What term is used to describe the temperature given as 18°C in this case?
(c)(i) Write an equation for the action of heat on each of the following compounds:
I. \( \mathrm{AgNO_3} \)
II. \( \mathrm{(NH4)_2CO_3} \).
(ii) Copy and complete the table below
| Metal | Name of main ore | Method of extraction | One major use Haematite |
| — | Haematite | — | — |
| — | — | Electrolysis of molten oxide | — |
(a) Two substances for each purpose:
| Purpose | Substances |
|---|---|
| (i) Dry hydrogen | Concentrated H2SO4; fused (anhydrous) calcium chloride, CaCl2 |
| (ii) Remove CO2 from air | Sodium hydroxide solution (NaOH); soda lime |
| (iii) Convert hot CuO to copper | Hydrogen gas; carbon (II) oxide, CO (coke/carbon also acceptable) |
| (iv) Prepare ZnCl2 with dilute HCl | Zinc metal; zinc carbonate, ZnCO3 (zinc oxide also acceptable) |
(b)(i) Two alloys that contain lead: solder (tin and lead) and type metal (lead, tin and antimony).
(b)(ii) On bubbling H2S into a solution of Pb(NO3)2, a black precipitate is observed. This is lead(II) sulfide, PbS, which is insoluble and black:
\[\text{Pb(NO}_3)_2(aq) + \text{H}_2\text{S}(g) \rightarrow \text{PbS}(s) + 2\text{HNO}_3(aq)\](b)(iii) Metal M is tin.
- I. The phenomenon (existing in two different solid forms) is allotropy (the change from white to grey tin is known as tin pest).
- II. The temperature of 18 °C, at which one form changes into the other, is the transition temperature.
(c)(i) Action of heat:
I. On silver trioxonitrate(V):
\[2\text{AgNO}_3 \xrightarrow{\ \Delta\ } 2\text{Ag} + 2\text{NO}_2 + \text{O}_2\]II. On ammonium trioxocarbonate(IV):
\[(\text{NH}_4)_2\text{CO}_3 \xrightarrow{\ \Delta\ } 2\text{NH}_3 + \text{H}_2\text{O} + \text{CO}_2\](c)(ii) Completed table:
| Metal | Name of main ore | Method of extraction | One major use |
|---|---|---|---|
| Iron | Haematite | Reduction in the blast furnace (with coke / carbon (II) oxide) | Making steel for construction |
| Aluminium | Bauxite | Electrolysis of molten oxide (alumina) | Making aircraft bodies and cooking utensils |
Answer Details
(a) Two substances for each purpose:
| Purpose | Substances |
|---|---|
| (i) Dry hydrogen | Concentrated H2SO4; fused (anhydrous) calcium chloride, CaCl2 |
| (ii) Remove CO2 from air | Sodium hydroxide solution (NaOH); soda lime |
| (iii) Convert hot CuO to copper | Hydrogen gas; carbon (II) oxide, CO (coke/carbon also acceptable) |
| (iv) Prepare ZnCl2 with dilute HCl | Zinc metal; zinc carbonate, ZnCO3 (zinc oxide also acceptable) |
(b)(i) Two alloys that contain lead: solder (tin and lead) and type metal (lead, tin and antimony).
(b)(ii) On bubbling H2S into a solution of Pb(NO3)2, a black precipitate is observed. This is lead(II) sulfide, PbS, which is insoluble and black:
\[\text{Pb(NO}_3)_2(aq) + \text{H}_2\text{S}(g) \rightarrow \text{PbS}(s) + 2\text{HNO}_3(aq)\](b)(iii) Metal M is tin.
- I. The phenomenon (existing in two different solid forms) is allotropy (the change from white to grey tin is known as tin pest).
- II. The temperature of 18 °C, at which one form changes into the other, is the transition temperature.
(c)(i) Action of heat:
I. On silver trioxonitrate(V):
\[2\text{AgNO}_3 \xrightarrow{\ \Delta\ } 2\text{Ag} + 2\text{NO}_2 + \text{O}_2\]II. On ammonium trioxocarbonate(IV):
\[(\text{NH}_4)_2\text{CO}_3 \xrightarrow{\ \Delta\ } 2\text{NH}_3 + \text{H}_2\text{O} + \text{CO}_2\](c)(ii) Completed table:
| Metal | Name of main ore | Method of extraction | One major use |
|---|---|---|---|
| Iron | Haematite | Reduction in the blast furnace (with coke / carbon (II) oxide) | Making steel for construction |
| Aluminium | Bauxite | Electrolysis of molten oxide (alumina) | Making aircraft bodies and cooking utensils |
Question 6 Report
(a) Describe briefly a suitable procedure for preparing a pure sample of MgSO\(_4\) starting from MgO.
(b)(i) Mention two sources of water pollution.
(ii) Explain why the sample of air collected in the process of boiling water is richer in oxygen than atmospheric air
(iii) Mention one substance used as coagulant in water treatment plants.
(c)(i) State two physical porperties of chlorine.
(ii) Write an equation to show how chlorine reacts with iron
(iii) Why is Chlorine preferred to sulphur (IV) oxide in the bleaching of cotton
(d) Bleaching powder reacts with dilute HCl according to the reaction below;
CaOCl\(_{2(s)}\) + 2HCI\(_{(aq)}\) -> CaCl\(_{2(aq)}\) + H\(_2\)O\(_{(l)}\) + Cl\(_{2(g)}\)
Calculate the mass of bleaching powder that will produce 400cm\(^3\) of chlorine at 25\(^o\)C and a pressure of 1.20 x 10\(^5\) NM\(^{-2}\). [O = 16.0; Cl = 35.5; Ca = 40.0;1 mole of gas occupies 22.4 dm\(^3\) at s.t.p; standard pressure = 1.01 x 10\(^6\) Nm\(^{-2}\)]
(a) Preparing pure MgSO4 from MgO
Warm some dilute tetraoxosulphate(VI) acid in a beaker and add magnesium oxide (a base) a little at a time, stirring, until no more dissolves (excess MgO ensures all the acid is used up). Filter to remove the unreacted MgO. Evaporate the filtrate until saturated (to the point of crystallisation), allow to cool so that MgSO4·7H2O crystallises, then filter off and dry the crystals between filter papers.
\[ MgO + H_2SO_4 \to MgSO_4 + H_2O \]
(b) Water
(i) Two sources of water pollution: industrial effluents/waste; sewage or domestic waste (also agricultural run-off, oil spillage).
(ii) Oxygen is more soluble in water than nitrogen, so the dissolved air has a higher proportion of oxygen than atmospheric air; when this air is expelled by boiling, it is richer in oxygen (about 33% O2) than ordinary air (about 21% O2).
(iii) Coagulant: alum (aluminium sulphate).
(c) Chlorine
(i) Two physical properties: it is a greenish-yellow gas with a choking, pungent smell (also: denser than air, moderately soluble in water, poisonous).
(ii) \[ 2Fe + 3Cl_2 \to 2FeCl_3 \]
(iii) Chlorine is preferred to sulphur(IV) oxide because chlorine bleaches by oxidation, giving a permanent effect, whereas SO2 bleaches by reduction, which is temporary (the colour returns on exposure to air).
(d) Mass of bleaching powder
\[ CaOCl_2 + 2HCl \to CaCl_2 + H_2O + Cl_2 \]
Convert 400 cm3 of Cl2 at 25°C (298 K) and \(1.20 \times 10^5\ \text{N m}^{-2}\) to s.t.p. (273 K, \(1.01 \times 10^5\ \text{N m}^{-2}\)) using \( \dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2} \):
\[ V_2 = \frac{1.20\times10^5 \times 0.400 \times 273}{298 \times 1.01\times10^5} = 0.435\ \text{dm}^3 \]
Moles of Cl2 \( = \dfrac{0.435}{22.4} = 1.94 \times 10^{-2}\ \text{mol} \)
From the equation, 1 mol CaOCl2 gives 1 mol Cl2, so moles of CaOCl2 \( = 1.94 \times 10^{-2}\ \text{mol} \).
Molar mass of CaOCl2 \( = 40 + 16 + 71 = 127\ \text{g mol}^{-1} \)
Mass \( = 1.94 \times 10^{-2} \times 127 = \mathbf{2.47\ g} \)
(The stated "standard pressure = 1.01 × 106" is taken as 1.01 × 105 N m-2, the true standard pressure.)
Answer Details
(a) Preparing pure MgSO4 from MgO
Warm some dilute tetraoxosulphate(VI) acid in a beaker and add magnesium oxide (a base) a little at a time, stirring, until no more dissolves (excess MgO ensures all the acid is used up). Filter to remove the unreacted MgO. Evaporate the filtrate until saturated (to the point of crystallisation), allow to cool so that MgSO4·7H2O crystallises, then filter off and dry the crystals between filter papers.
\[ MgO + H_2SO_4 \to MgSO_4 + H_2O \]
(b) Water
(i) Two sources of water pollution: industrial effluents/waste; sewage or domestic waste (also agricultural run-off, oil spillage).
(ii) Oxygen is more soluble in water than nitrogen, so the dissolved air has a higher proportion of oxygen than atmospheric air; when this air is expelled by boiling, it is richer in oxygen (about 33% O2) than ordinary air (about 21% O2).
(iii) Coagulant: alum (aluminium sulphate).
(c) Chlorine
(i) Two physical properties: it is a greenish-yellow gas with a choking, pungent smell (also: denser than air, moderately soluble in water, poisonous).
(ii) \[ 2Fe + 3Cl_2 \to 2FeCl_3 \]
(iii) Chlorine is preferred to sulphur(IV) oxide because chlorine bleaches by oxidation, giving a permanent effect, whereas SO2 bleaches by reduction, which is temporary (the colour returns on exposure to air).
(d) Mass of bleaching powder
\[ CaOCl_2 + 2HCl \to CaCl_2 + H_2O + Cl_2 \]
Convert 400 cm3 of Cl2 at 25°C (298 K) and \(1.20 \times 10^5\ \text{N m}^{-2}\) to s.t.p. (273 K, \(1.01 \times 10^5\ \text{N m}^{-2}\)) using \( \dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2} \):
\[ V_2 = \frac{1.20\times10^5 \times 0.400 \times 273}{298 \times 1.01\times10^5} = 0.435\ \text{dm}^3 \]
Moles of Cl2 \( = \dfrac{0.435}{22.4} = 1.94 \times 10^{-2}\ \text{mol} \)
From the equation, 1 mol CaOCl2 gives 1 mol Cl2, so moles of CaOCl2 \( = 1.94 \times 10^{-2}\ \text{mol} \).
Molar mass of CaOCl2 \( = 40 + 16 + 71 = 127\ \text{g mol}^{-1} \)
Mass \( = 1.94 \times 10^{-2} \times 127 = \mathbf{2.47\ g} \)
(The stated "standard pressure = 1.01 × 106" is taken as 1.01 × 105 N m-2, the true standard pressure.)
