WAEC SSCE - Chemistry - 2000 (Essay)
Tambaya 1 Rahoto
(a) What is meant by each of the following terms?:
(i) Esterification
(ii) Saponification
(b)(i) Give the general moluecular 1 formula of alkynes
(ii) Write the molecular formula and empirical formula of ethylethanoate.
(iii) Draw the structure of 1, 1, 2, 2-tetrabromoethane
(iv) Write an equation for the reaction of ethanol with sodium
(c) Consider the following reaction schemes:
| I | II | |||
| Petroleum | ---> | Petroleum Fractions. Higher Petroleum Fractions | ---> | Petrol + X |
(i) State type of process/reaction involved in each of the stages labelled I to IV.
(ii) Identify X and Y
(iii) Give the IUPAC name of the product obtained in stage III.
(iv) What are the reaction conditions for stage IV?
(d) Explain why palm wine: (i) froths or foams (ii) tastes sour after some days.
(a)
(i) Esterification is the reaction between an alkanol and an alkanoic acid, in the presence of a mineral acid catalyst such as concentrated sulphuric acid, to form an ester and water.
(ii) Saponification is the alkaline hydrolysis of a fat, oil or ester with an alkali such as sodium hydroxide to produce soap, which is the sodium salt of a fatty acid, and glycerol in the case of fats and oils.
(b)
(i) The general molecular formula of alkynes is: \[\mathrm{C_nH_{2n-2}}\]
(ii) Ethyl ethanoate has molecular formula \(\mathrm{C_4H_8O_2}\). Its empirical formula is \(\mathrm{C_2H_4O}\).
(iii) The displayed structure of 1,1,2,2-tetrabromoethane is:
(iv)
\[2\mathrm{C_2H_5OH}+2\mathrm{Na}\rightarrow2\mathrm{C_2H_5ONa}+\mathrm{H_2}\]
(c)
(i)
- Fractional distillation.
- Cracking, usually catalytic cracking.
- Addition reaction, specifically chlorination of ethene.
- Addition polymerisation.
(ii) \(X\) is ethene, \(\mathrm{C_2H_4}\), and \(Y\) is chlorine, \(\mathrm{Cl_2}\).
(iii) The IUPAC name of the product in stage III is 1,2-dichloroethane.
\[\mathrm{CH_2{=}CH_2}+\mathrm{Cl_2}\rightarrow\mathrm{ClCH_2CH_2Cl}\]
(iv) Ethene is polymerised at a high temperature of about \(150\text{ to }300^\circ\mathrm{C}\) and high pressure of about \(500\text{ to }1500\) atmospheres, using a trace of oxygen or an organic peroxide as initiator.
(d)
(i) Palm wine froths because yeast ferments the sugars present to ethanol and carbon dioxide. The bubbles of carbon dioxide cause the foaming.
\[\mathrm{C_6H_{12}O_6}\xrightarrow{\text{yeast}}2\mathrm{C_2H_5OH}+2\mathrm{CO_2}\]
(ii) After some days, bacteria oxidise the ethanol in palm wine, in the presence of atmospheric oxygen, to ethanoic acid. Ethanoic acid gives the wine its sour taste.
\[\mathrm{C_2H_5OH}+\mathrm{O_2}\xrightarrow{\text{bacteria}}\mathrm{CH_3COOH}+\mathrm{H_2O}\]
Bayanin Amsa
(a)
(i) Esterification is the reaction between an alkanol and an alkanoic acid, in the presence of a mineral acid catalyst such as concentrated sulphuric acid, to form an ester and water.
(ii) Saponification is the alkaline hydrolysis of a fat, oil or ester with an alkali such as sodium hydroxide to produce soap, which is the sodium salt of a fatty acid, and glycerol in the case of fats and oils.
(b)
(i) The general molecular formula of alkynes is: \[\mathrm{C_nH_{2n-2}}\]
(ii) Ethyl ethanoate has molecular formula \(\mathrm{C_4H_8O_2}\). Its empirical formula is \(\mathrm{C_2H_4O}\).
(iii) The displayed structure of 1,1,2,2-tetrabromoethane is:
(iv)
\[2\mathrm{C_2H_5OH}+2\mathrm{Na}\rightarrow2\mathrm{C_2H_5ONa}+\mathrm{H_2}\]
(c)
(i)
- Fractional distillation.
- Cracking, usually catalytic cracking.
- Addition reaction, specifically chlorination of ethene.
- Addition polymerisation.
(ii) \(X\) is ethene, \(\mathrm{C_2H_4}\), and \(Y\) is chlorine, \(\mathrm{Cl_2}\).
(iii) The IUPAC name of the product in stage III is 1,2-dichloroethane.
\[\mathrm{CH_2{=}CH_2}+\mathrm{Cl_2}\rightarrow\mathrm{ClCH_2CH_2Cl}\]
(iv) Ethene is polymerised at a high temperature of about \(150\text{ to }300^\circ\mathrm{C}\) and high pressure of about \(500\text{ to }1500\) atmospheres, using a trace of oxygen or an organic peroxide as initiator.
(d)
(i) Palm wine froths because yeast ferments the sugars present to ethanol and carbon dioxide. The bubbles of carbon dioxide cause the foaming.
\[\mathrm{C_6H_{12}O_6}\xrightarrow{\text{yeast}}2\mathrm{C_2H_5OH}+2\mathrm{CO_2}\]
(ii) After some days, bacteria oxidise the ethanol in palm wine, in the presence of atmospheric oxygen, to ethanoic acid. Ethanoic acid gives the wine its sour taste.
\[\mathrm{C_2H_5OH}+\mathrm{O_2}\xrightarrow{\text{bacteria}}\mathrm{CH_3COOH}+\mathrm{H_2O}\]
Tambaya 2 Rahoto
a)(i) State Graham's law of diffusion.
(ii) Calculate the vapour density of a triatomic gas X if its relative: atomic mass is 16.
(iii) Equal volumes of gases Y and Z are maintained at the same temperature and pressure. If the mass of a molecule of Y is twice that of Z state and explain which of the molecules has the, greater average velocity.
(b) The graph below is the ratio curve for the following reaction carried out in an open vessel.
MgCO\(_{3(s)}\) + 2HCI\(_{(aq)}\) \(\to\) MgCl\(_{2(aq)}\) + CO\(_{2(g)}\) + H\(_2\)\(_{(l)}\).
(i) For how long did reaction occur?
(ii) Why was there a loss in mass?
(iii) State whether reaction rate was fastest at the beginning, the middle or towards the end of the reaction. Give reason for our answer.
(iv) List three reaction conditions that can affect the slope of the curve
(c) Consider the following reaction at equilibrium: PCI\(_{5(g)}\) \(\rightleftharpoons\) PCI\(_{3(g)}\)); \(\Delta\)H = +95 kJmol\(^{-}\)
(i) Write an expression for the equilibrium constant K.
(ii) Predict the effect of the following on the equilibrium position.
I. Increased pressure
II. Increased temperature
III. Removal of chlorine Sketch an energy profile diagram for the forward reaction.
(a)(i) Graham's law of diffusion states that, at constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its density or relative molecular mass.
\[r \propto \frac{1}{\sqrt{d}}\qquad \text{or}\qquad r \propto \frac{1}{\sqrt{M}}\]
(ii) Since X is triatomic, its formula is \(X_3\).
\[M_r(X_3)=3\times16=48\]
\[\text{Vapour density}=\frac{M_r}{2}=\frac{48}{2}=24\]
Vapour density of X = 24.
(iii) Molecules of Z have the greater average velocity. At the same temperature, molecules of Y and Z have the same average kinetic energy. Since a molecule of Y has twice the mass of a molecule of Z, the lighter Z molecule moves faster.
\[\frac{1}{2}m_Yv_Y^2=\frac{1}{2}m_Zv_Z^2\]
Since \(m_Y=2m_Z\),
\[v_Z=\sqrt{2}\,v_Y\]
(b)(i) The reaction occurred for 12 minutes, after which the curve became horizontal.
(ii) Carbon(IV) oxide, \(\mathrm{CO_2}\), was evolved and escaped from the open vessel. This caused the mass of the flask and its contents to decrease.
(iii) The reaction rate was fastest at the beginning of the reaction because the curve had its greatest gradient at the beginning. The acid concentration was then highest, so effective collisions occurred most frequently.
(iv) Conditions that affect the slope of the curve include:
- concentration of hydrochloric acid;
- temperature of the reaction mixture;
- surface area or particle size of magnesium carbonate.
(c) \[\mathrm{PCl_{5(g)}\rightleftharpoons PCl_{3(g)}+Cl_{2(g)}}\qquad \Delta H=+95\ \mathrm{kJ\,mol^{-1}}\]
(i)
\[K_c=\frac{[\mathrm{PCl_3}][\mathrm{Cl_2}]}{[\mathrm{PCl_5}]}\]
(ii)
- Increased pressure: equilibrium shifts to the left, favouring \(\mathrm{PCl_5}\), because there is one mole of gas on the left but two moles of gas on the right.
- Increased temperature: equilibrium shifts to the right, because the forward reaction is endothermic.
- Removal of chlorine: equilibrium shifts to the right to replace the removed \(\mathrm{Cl_2}\).
Energy profile diagram for the forward reaction
The products are at a higher energy level than the reactant; hence \(\Delta H\) is positive and the forward reaction is endothermic.
Bayanin Amsa
(a)(i) Graham's law of diffusion states that, at constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its density or relative molecular mass.
\[r \propto \frac{1}{\sqrt{d}}\qquad \text{or}\qquad r \propto \frac{1}{\sqrt{M}}\]
(ii) Since X is triatomic, its formula is \(X_3\).
\[M_r(X_3)=3\times16=48\]
\[\text{Vapour density}=\frac{M_r}{2}=\frac{48}{2}=24\]
Vapour density of X = 24.
(iii) Molecules of Z have the greater average velocity. At the same temperature, molecules of Y and Z have the same average kinetic energy. Since a molecule of Y has twice the mass of a molecule of Z, the lighter Z molecule moves faster.
\[\frac{1}{2}m_Yv_Y^2=\frac{1}{2}m_Zv_Z^2\]
Since \(m_Y=2m_Z\),
\[v_Z=\sqrt{2}\,v_Y\]
(b)(i) The reaction occurred for 12 minutes, after which the curve became horizontal.
(ii) Carbon(IV) oxide, \(\mathrm{CO_2}\), was evolved and escaped from the open vessel. This caused the mass of the flask and its contents to decrease.
(iii) The reaction rate was fastest at the beginning of the reaction because the curve had its greatest gradient at the beginning. The acid concentration was then highest, so effective collisions occurred most frequently.
(iv) Conditions that affect the slope of the curve include:
- concentration of hydrochloric acid;
- temperature of the reaction mixture;
- surface area or particle size of magnesium carbonate.
(c) \[\mathrm{PCl_{5(g)}\rightleftharpoons PCl_{3(g)}+Cl_{2(g)}}\qquad \Delta H=+95\ \mathrm{kJ\,mol^{-1}}\]
(i)
\[K_c=\frac{[\mathrm{PCl_3}][\mathrm{Cl_2}]}{[\mathrm{PCl_5}]}\]
(ii)
- Increased pressure: equilibrium shifts to the left, favouring \(\mathrm{PCl_5}\), because there is one mole of gas on the left but two moles of gas on the right.
- Increased temperature: equilibrium shifts to the right, because the forward reaction is endothermic.
- Removal of chlorine: equilibrium shifts to the right to replace the removed \(\mathrm{Cl_2}\).
Energy profile diagram for the forward reaction
The products are at a higher energy level than the reactant; hence \(\Delta H\) is positive and the forward reaction is endothermic.
Tambaya 3 Rahoto
(a)(i) List three properties of elements which increase generally across a period in the Periodic Table,
(ii) Give two differences between a chemical reaction and a nuclear reaction.
(b) Use the information provided in the table below to answer Questions (i) - (vii).
| Atom of Element | P | Q | R | S | T |
| Mass Number | 16 | 40 | 35 | 18 | 20 |
| Atomic Number | 8 | 20 | 17 | 8 | 10 |
Which of the atoms in the table above:
(i) are isotopes of the same element?;
(ii) contains 18 neutrons?;
(iii) is chemically unreactive?;
(iv) readily forms an ion with two positive charges?
(v) attain an octet structure by accepting one electron?;
(vi) forms ionic bond with R?;
(vii) belongs to the s-block in the Periodic Table?
(c) Describe in outline how each of the following conversions can be carried out in the laboratory. Write appropriate equations for the reactions involved in each case
(i) \( \mathrm{CuCO_3} \) to \( \mathrm{Cu} \)
(ii) \( \mathrm{MgO} \) to \( \mathrm{MgSO_4} \).
(a)(i) Three properties that increase generally across a period: ionisation energy, electronegativity, and electron affinity (non-metallic character also increases).
(a)(ii) Two differences between a chemical reaction and a nuclear reaction:
| Chemical reaction | Nuclear reaction |
|---|---|
| Involves only the outer (valence) electrons; the nucleus is unchanged. | Involves the nucleus itself (protons and neutrons change). |
| Relatively small energy changes; no new element is formed. | Enormous energy changes; new elements are formed (transmutation). |
(b) The atoms
| Element | P | Q | R | S | T |
|---|---|---|---|---|---|
| Mass number | 16 | 40 | 35 | 18 | 20 |
| Atomic number | 8 | 20 | 17 | 8 | 10 |
| Neutrons (mass - atomic no.) | 8 | 20 | 18 | 10 | 10 |
- (i) Isotopes of the same element: P and S (both have atomic number 8).
- (ii) Contains 18 neutrons: R (35 - 17 = 18).
- (iii) Chemically unreactive: T (atomic number 10, a noble gas with a stable octet).
- (iv) Readily forms an ion with two positive charges: Q (atomic number 20, group II, forms Q\(^{2+}\)).
- (v) Attains an octet by accepting one electron: R (atomic number 17, seven outer electrons, gains one).
- (vi) Forms an ionic bond with R: Q (a metal that donates its two outer electrons to the non-metal R).
- (vii) Belongs to the s-block: Q (its outer electrons occupy the 4s sub-shell).
(c)(i) CuCO\(_3\) to Cu
Heat the copper(II) trioxocarbonate(IV) strongly to give the oxide, then reduce the oxide by passing hydrogen (or heating with carbon) over it:
\[ \text{CuCO}_3 \xrightarrow{\text{heat}} \text{CuO} + \text{CO}_2 \] \[ \text{CuO} + \text{H}_2 \xrightarrow{\text{heat}} \text{Cu} + \text{H}_2\text{O} \](c)(ii) MgO to MgSO\(_4\)
Add the magnesium oxide (a base) to warm dilute tetraoxosulphate(VI) acid until no more dissolves; filter off excess oxide, then evaporate and crystallise the filtrate:
\[ \text{MgO} + \text{H}_2\text{SO}_4 \rightarrow \text{MgSO}_4 + \text{H}_2\text{O} \]Bayanin Amsa
(a)(i) Three properties that increase generally across a period: ionisation energy, electronegativity, and electron affinity (non-metallic character also increases).
(a)(ii) Two differences between a chemical reaction and a nuclear reaction:
| Chemical reaction | Nuclear reaction |
|---|---|
| Involves only the outer (valence) electrons; the nucleus is unchanged. | Involves the nucleus itself (protons and neutrons change). |
| Relatively small energy changes; no new element is formed. | Enormous energy changes; new elements are formed (transmutation). |
(b) The atoms
| Element | P | Q | R | S | T |
|---|---|---|---|---|---|
| Mass number | 16 | 40 | 35 | 18 | 20 |
| Atomic number | 8 | 20 | 17 | 8 | 10 |
| Neutrons (mass - atomic no.) | 8 | 20 | 18 | 10 | 10 |
- (i) Isotopes of the same element: P and S (both have atomic number 8).
- (ii) Contains 18 neutrons: R (35 - 17 = 18).
- (iii) Chemically unreactive: T (atomic number 10, a noble gas with a stable octet).
- (iv) Readily forms an ion with two positive charges: Q (atomic number 20, group II, forms Q\(^{2+}\)).
- (v) Attains an octet by accepting one electron: R (atomic number 17, seven outer electrons, gains one).
- (vi) Forms an ionic bond with R: Q (a metal that donates its two outer electrons to the non-metal R).
- (vii) Belongs to the s-block: Q (its outer electrons occupy the 4s sub-shell).
(c)(i) CuCO\(_3\) to Cu
Heat the copper(II) trioxocarbonate(IV) strongly to give the oxide, then reduce the oxide by passing hydrogen (or heating with carbon) over it:
\[ \text{CuCO}_3 \xrightarrow{\text{heat}} \text{CuO} + \text{CO}_2 \] \[ \text{CuO} + \text{H}_2 \xrightarrow{\text{heat}} \text{Cu} + \text{H}_2\text{O} \](c)(ii) MgO to MgSO\(_4\)
Add the magnesium oxide (a base) to warm dilute tetraoxosulphate(VI) acid until no more dissolves; filter off excess oxide, then evaporate and crystallise the filtrate:
\[ \text{MgO} + \text{H}_2\text{SO}_4 \rightarrow \text{MgSO}_4 + \text{H}_2\text{O} \]Tambaya 4 Rahoto
(a)(i) Give the names of two allotropes of sulphur.
(ii) State and explain what is observed when hydrogen sulphide is bubbled through acidified potassium tetraoxomanganatc (VII) solution
(iii) List one product of the reaction of sulphur (IV) oxide with hydrogen sulphide
(b)(i) What are the raw materials for the manufacture of tetraoxosulphate (VI) acid by the contact process?
(ii) Write an equation for the reaction that requires a catalyst in the contact process and state the catalyst used.
(iii) State the observation and the product formed when concentrated \(H_2SO_4\) reacts with each of the following:
I. Copper turnings Ii. A cube of sugar
(c)(i) Give three uses of sodium trioxocarbonate (IV).
(ii) What name is given to reactions of the following type?:
\[ \mathrm{Na_2CO_3 + 10H_2O_{(s)} \xrightarrow{\text{exposure to air}} Na_2CO_3; H_2O_{(S)} + 9H_2O} \](iii) Calculate the solubility of \(Na_2CO_3\) at 25°C, if 2.0 cm\(^3\) of its saturated solution at that temperature gave 1.75g of the anhydrous alt. [C = 12, O = 16, Na = 23].
(a)(i) Two allotropes of sulphur: rhombic (octahedral/alpha) sulphur and monoclinic (prismatic/beta) sulphur.
(a)(ii) When H2S is bubbled through acidified KMnO4, the purple solution is decolourised and a pale yellow precipitate of sulphur forms. H2S is a reducing agent: it reduces Mn7+ (purple) to colourless Mn2+ and is itself oxidised to sulphur.
(a)(iii) Product of SO2 with H2S: sulphur (and water): \(\text{SO}_2 + 2\text{H}_2\text{S} \to 3\text{S} + 2\text{H}_2\text{O}\).
(b)(i) Raw materials for the contact process: sulphur (or a sulphide ore/roasted iron pyrites), air (oxygen) and water.
(b)(ii) Catalysed step: \(2\text{SO}_2 + \text{O}_2 \rightleftharpoons 2\text{SO}_3\); catalyst: vanadium(V) oxide, V2O5.
(b)(iii) Concentrated H2SO4 with:
- I. Copper turnings (hot): the acid acts as an oxidizing agent; the mixture turns blue, SO2 is evolved and copper(II) tetraoxosulphate(VI) forms: \(\text{Cu} + 2\text{H}_2\text{SO}_4 \to \text{CuSO}_4 + \text{SO}_2 + 2\text{H}_2\text{O}\).
- II. A cube of sugar: the acid acts as a dehydrating agent; the sugar chars to a black spongy mass of carbon: \(\text{C}_{12}\text{H}_{22}\text{O}_{11} \to 12\text{C} + 11\text{H}_2\text{O}\).
(c)(i) Uses of sodium trioxocarbonate(IV) (Na2CO3): in the manufacture of glass; softening of hard water; manufacture of soap and detergents; as washing soda.
(c)(ii) The loss of water of crystallisation on exposure to air is called efflorescence.
(c)(iii) Solubility of Na2CO3 at 25 °C: molar mass Na2CO3 \(= 106\ \text{g mol}^{-1}\).
\[ \text{moles in }2.0\ \text{cm}^3 = \frac{1.75}{106} = 0.0165\ \text{mol} \] \[ \text{solubility} = \frac{0.0165}{2.0/1000} = 8.25\ \text{mol dm}^{-3} \]Bayanin Amsa
(a)(i) Two allotropes of sulphur: rhombic (octahedral/alpha) sulphur and monoclinic (prismatic/beta) sulphur.
(a)(ii) When H2S is bubbled through acidified KMnO4, the purple solution is decolourised and a pale yellow precipitate of sulphur forms. H2S is a reducing agent: it reduces Mn7+ (purple) to colourless Mn2+ and is itself oxidised to sulphur.
(a)(iii) Product of SO2 with H2S: sulphur (and water): \(\text{SO}_2 + 2\text{H}_2\text{S} \to 3\text{S} + 2\text{H}_2\text{O}\).
(b)(i) Raw materials for the contact process: sulphur (or a sulphide ore/roasted iron pyrites), air (oxygen) and water.
(b)(ii) Catalysed step: \(2\text{SO}_2 + \text{O}_2 \rightleftharpoons 2\text{SO}_3\); catalyst: vanadium(V) oxide, V2O5.
(b)(iii) Concentrated H2SO4 with:
- I. Copper turnings (hot): the acid acts as an oxidizing agent; the mixture turns blue, SO2 is evolved and copper(II) tetraoxosulphate(VI) forms: \(\text{Cu} + 2\text{H}_2\text{SO}_4 \to \text{CuSO}_4 + \text{SO}_2 + 2\text{H}_2\text{O}\).
- II. A cube of sugar: the acid acts as a dehydrating agent; the sugar chars to a black spongy mass of carbon: \(\text{C}_{12}\text{H}_{22}\text{O}_{11} \to 12\text{C} + 11\text{H}_2\text{O}\).
(c)(i) Uses of sodium trioxocarbonate(IV) (Na2CO3): in the manufacture of glass; softening of hard water; manufacture of soap and detergents; as washing soda.
(c)(ii) The loss of water of crystallisation on exposure to air is called efflorescence.
(c)(iii) Solubility of Na2CO3 at 25 °C: molar mass Na2CO3 \(= 106\ \text{g mol}^{-1}\).
\[ \text{moles in }2.0\ \text{cm}^3 = \frac{1.75}{106} = 0.0165\ \text{mol} \] \[ \text{solubility} = \frac{0.0165}{2.0/1000} = 8.25\ \text{mol dm}^{-3} \]Tambaya 5 Rahoto
(a) State how you would carry out the following procedures in the laboratory:
(i) Remove the sediment in sample of water;
(ii) Soften temporarily hard water without heating it,
(iii) Obtain pure water frdm muddy water;
(iv) Remove oxygen and moisture from a sample of air
(b)(i) What type of salts are alums?
(ii) State the function of alum in water treatment plants.
(iii) State and explain how rain water that hac passed through limestone deposits will react with soap solution.
(c)(i) Write an equation for the laboraton preparation of chlorine
(ii) List the products of the reaction of chorine with hot concentrated sodiuri hydroxide solution
(iii) What is observed when moist blue litmus paper comes in contact with chlorine?
(iv) Calculate the volume of chlorine at s.t.p. that would be required to react completely with 3.70g of dry slaked lime according to the following equation:
Ca(OH)\(_{2(s)}\) + Cl\(_{2(g)}\) --> CaOCl\(_{2}\). H\(_2\)O\(_{(s)}\) [H = 1, O = 16, Ca = 40, 1 mole of gas occupies 22.4 dm\(^3\) at s.t.p.]
(d)(i) State what is observed on warming ammonium trioxonitrate (V) with sodium hydroxide solution
(ii) Explain why ammonium trioxocarbonate (IV) leaves no residue on being heated.
(a) Laboratory procedures:
- (i) Remove sediment from water: by filtration (or decantation/sedimentation followed by filtration).
- (ii) Soften temporary hard water without heating: add a calculated amount of calcium hydroxide (slaked lime) - Clark's method - or add sodium trioxocarbonate(IV) (washing soda) to precipitate the Ca2+/Mg2+ ions.
- (iii) Obtain pure water from muddy water: by (simple) distillation (filter first, then distil).
- (iv) Remove oxygen and moisture from air: pass the air over heated copper to remove oxygen, then through a drying agent such as concentrated H2SO4 (or fused CaCl2) to remove moisture.
(b) (i) Alums are double salts. (ii) In water treatment, alum acts as a coagulant/flocculant: it causes the fine suspended clay/colloidal particles to clump together and settle out, clarifying the water. (iii) Rain water that has passed through limestone deposits dissolves calcium hydrogentrioxocarbonate(IV), becoming temporary hard water; with soap it gives little lather and forms an insoluble scum (because Ca2+ reacts with the soap) before it can lather.
(c) Chlorine
(i) Laboratory preparation: \(\text{MnO}_2 + 4\text{HCl} \to \text{MnCl}_2 + \text{Cl}_2 + 2\text{H}_2\text{O}\) (conc. HCl, warmed).
(ii) Products of chlorine with hot concentrated NaOH: sodium chloride, sodium trioxochlorate(V) (chlorate, NaClO3) and water: \(3\text{Cl}_2 + 6\text{NaOH} \to 5\text{NaCl} + \text{NaClO}_3 + 3\text{H}_2\text{O}\).
(iii) Moist blue litmus paper turns red and is then bleached (decolourised) by the chlorine.
(iv) Volume of Cl2 at s.t.p. to react with 3.70 g of slaked lime, Ca(OH)2 (molar mass = 74):
\[ \text{moles Ca(OH)}_2 = \frac{3.70}{74} = 0.05\ \text{mol} \]The ratio Ca(OH)2 : Cl2 is 1 : 1, so moles Cl2 = 0.05 mol.
\[ \text{Volume} = 0.05 \times 22.4 = 1.12\ \text{dm}^3 \](d) (i) Warming ammonium trioxonitrate(V) with NaOH: a pungent gas (ammonia) is evolved that turns moist red litmus blue: \(\text{NH}_4\text{NO}_3 + \text{NaOH} \to \text{NaNO}_3 + \text{H}_2\text{O} + \text{NH}_3\). (ii) Ammonium trioxocarbonate(IV) leaves no residue on heating because it decomposes completely into gaseous products only (ammonia, carbon(IV) oxide and steam): \((\text{NH}_4)_2\text{CO}_3 \to 2\text{NH}_3 + \text{CO}_2 + \text{H}_2\text{O}\), all of which escape.
Bayanin Amsa
(a) Laboratory procedures:
- (i) Remove sediment from water: by filtration (or decantation/sedimentation followed by filtration).
- (ii) Soften temporary hard water without heating: add a calculated amount of calcium hydroxide (slaked lime) - Clark's method - or add sodium trioxocarbonate(IV) (washing soda) to precipitate the Ca2+/Mg2+ ions.
- (iii) Obtain pure water from muddy water: by (simple) distillation (filter first, then distil).
- (iv) Remove oxygen and moisture from air: pass the air over heated copper to remove oxygen, then through a drying agent such as concentrated H2SO4 (or fused CaCl2) to remove moisture.
(b) (i) Alums are double salts. (ii) In water treatment, alum acts as a coagulant/flocculant: it causes the fine suspended clay/colloidal particles to clump together and settle out, clarifying the water. (iii) Rain water that has passed through limestone deposits dissolves calcium hydrogentrioxocarbonate(IV), becoming temporary hard water; with soap it gives little lather and forms an insoluble scum (because Ca2+ reacts with the soap) before it can lather.
(c) Chlorine
(i) Laboratory preparation: \(\text{MnO}_2 + 4\text{HCl} \to \text{MnCl}_2 + \text{Cl}_2 + 2\text{H}_2\text{O}\) (conc. HCl, warmed).
(ii) Products of chlorine with hot concentrated NaOH: sodium chloride, sodium trioxochlorate(V) (chlorate, NaClO3) and water: \(3\text{Cl}_2 + 6\text{NaOH} \to 5\text{NaCl} + \text{NaClO}_3 + 3\text{H}_2\text{O}\).
(iii) Moist blue litmus paper turns red and is then bleached (decolourised) by the chlorine.
(iv) Volume of Cl2 at s.t.p. to react with 3.70 g of slaked lime, Ca(OH)2 (molar mass = 74):
\[ \text{moles Ca(OH)}_2 = \frac{3.70}{74} = 0.05\ \text{mol} \]The ratio Ca(OH)2 : Cl2 is 1 : 1, so moles Cl2 = 0.05 mol.
\[ \text{Volume} = 0.05 \times 22.4 = 1.12\ \text{dm}^3 \](d) (i) Warming ammonium trioxonitrate(V) with NaOH: a pungent gas (ammonia) is evolved that turns moist red litmus blue: \(\text{NH}_4\text{NO}_3 + \text{NaOH} \to \text{NaNO}_3 + \text{H}_2\text{O} + \text{NH}_3\). (ii) Ammonium trioxocarbonate(IV) leaves no residue on heating because it decomposes completely into gaseous products only (ammonia, carbon(IV) oxide and steam): \((\text{NH}_4)_2\text{CO}_3 \to 2\text{NH}_3 + \text{CO}_2 + \text{H}_2\text{O}\), all of which escape.
Tambaya 6 Rahoto
(a)(i) List two properties of iron that are characteristic of transition metals
(ii) Using equations only, show the processes involved in the extraction of iron and the removal of impurities in the blast furnace.
(iii) The following reaction occurs when a piece of iron is exposed to moist air for some days:
4Fe\(_{(s)}\) + 3O\(_{2(g)}\) + xH\(_2\)O\(_{(l)}\) —> 2Fe\(_2\)O\(_3\)\(_{(s)}\) State three methods by which this reaction can be prevented
(iv) What is the oxidation number of iiron in the product in (iii) above?
(b)(i) Arrange the following metals in the order of increasing reactivity. Hence, state which of them is/are extracted by electrolysis Au, Zn, Mg, Na, Sn, Ca
(ii) Why is zinc said to be amphoteric?
(c)(i) Define oxidation in term of electron transfer
(ii) Determine how many moles of electrons are transferred when 4825 coulombs of electricity are passed through an electrolytic cell. [1F = 96500C]
(iii) Calculate the number of copper (II) ions that will be discharged by 0.250F. [Avogadro constant = 5.02 x 10\(^{23}\)
(a)(i) Two transition-metal properties of iron: it exhibits variable oxidation states (Fe2+ and Fe3+) and it forms coloured compounds (it also acts as a catalyst, e.g. in the Haber process).
(a)(ii) Extraction of iron in the blast furnace (equations):
\[ \text{C} + \text{O}_2 \to \text{CO}_2 \] \[ \text{CO}_2 + \text{C} \to 2\text{CO} \] \[ \text{Fe}_2\text{O}_3 + 3\text{CO} \to 2\text{Fe} + 3\text{CO}_2 \]Removal of impurities (limestone decomposes and the lime removes the silica gangue as slag):
\[ \text{CaCO}_3 \to \text{CaO} + \text{CO}_2 \] \[ \text{CaO} + \text{SiO}_2 \to \text{CaSiO}_3\ (\text{slag}) \](a)(iii) Prevention of rusting (three methods): painting the surface; coating with oil/grease; galvanising (coating with zinc) or electroplating; sacrificial protection with a more reactive metal.
(a)(iv) The product is Fe2O3; the oxidation number of iron in it is +3.
(b)(i) Increasing reactivity: Au < Sn < Zn < Mg < Ca < Na. Those extracted by electrolysis are the most reactive: Na, Ca and Mg.
(b)(ii) Zinc is amphoteric because it (and its oxide/hydroxide) reacts with both acids and alkalis to form a salt and water.
(c)(i) Oxidation: the loss of electrons by a species.
(c)(ii) Moles of electrons \(= \dfrac{Q}{F} = \dfrac{4825}{96500} = 0.05\ \text{mol}\).
(c)(iii) For \(\text{Cu}^{2+} + 2e^- \to \text{Cu}\), 0.250 F provides 0.250 mol of electrons, discharging \(\dfrac{0.250}{2} = 0.125\ \text{mol}\) of Cu2+.
\[ \text{Number of ions} = 0.125 \times 6.02\times10^{23} = 7.53\times10^{22}\ \text{ions} \]Bayanin Amsa
(a)(i) Two transition-metal properties of iron: it exhibits variable oxidation states (Fe2+ and Fe3+) and it forms coloured compounds (it also acts as a catalyst, e.g. in the Haber process).
(a)(ii) Extraction of iron in the blast furnace (equations):
\[ \text{C} + \text{O}_2 \to \text{CO}_2 \] \[ \text{CO}_2 + \text{C} \to 2\text{CO} \] \[ \text{Fe}_2\text{O}_3 + 3\text{CO} \to 2\text{Fe} + 3\text{CO}_2 \]Removal of impurities (limestone decomposes and the lime removes the silica gangue as slag):
\[ \text{CaCO}_3 \to \text{CaO} + \text{CO}_2 \] \[ \text{CaO} + \text{SiO}_2 \to \text{CaSiO}_3\ (\text{slag}) \](a)(iii) Prevention of rusting (three methods): painting the surface; coating with oil/grease; galvanising (coating with zinc) or electroplating; sacrificial protection with a more reactive metal.
(a)(iv) The product is Fe2O3; the oxidation number of iron in it is +3.
(b)(i) Increasing reactivity: Au < Sn < Zn < Mg < Ca < Na. Those extracted by electrolysis are the most reactive: Na, Ca and Mg.
(b)(ii) Zinc is amphoteric because it (and its oxide/hydroxide) reacts with both acids and alkalis to form a salt and water.
(c)(i) Oxidation: the loss of electrons by a species.
(c)(ii) Moles of electrons \(= \dfrac{Q}{F} = \dfrac{4825}{96500} = 0.05\ \text{mol}\).
(c)(iii) For \(\text{Cu}^{2+} + 2e^- \to \text{Cu}\), 0.250 F provides 0.250 mol of electrons, discharging \(\dfrac{0.250}{2} = 0.125\ \text{mol}\) of Cu2+.
\[ \text{Number of ions} = 0.125 \times 6.02\times10^{23} = 7.53\times10^{22}\ \text{ions} \]
