Loading....
|
Press & Hold to Drag Around |
|||
|
Click Here to Close |
|||
Question 1 Report
(a)(i) Draw and label a diagram to illustrate the preparation and collection of dry chlorine gas in the laboratory.
(ii) List two uses of chlorine.
(b)(i) Explain why river water flowing through an industrial town may be unsafe for drinking.
(ii) State the use of each of the following substances in water treatment: I. Sand, II. Chlorine, III. Calcium oxide, IV. Alum
(c)Consider the reaction represented by the following equation:
2Na\(_2\)CI\(_{(s)}\) + H\(_2\)SO\(_{4(aq)}\) \(\to\) Na\(_2\)SO\(_{4(aq)}\) + 2HCI\(_{(g)}\)
Calculate the volume of HCI gas that can be obtained at s.t.p. from 5.85 g of sodium chloride. [H = 1, Na = 23, CI = 35.5, Molar volume a 22.4 dm\(^3\) at s.t.p]
(d) Give one example in each case of a (i) metal that is a liquid at room temperature. (ii) non-metal that is a iiquid at room temperature, (iii) gas at room temperature that is monatomic.
(e) State two differences between metals and nom metals with respect to their: (i) physical properties; (ii) chemical properties.
(a)(i) Laboratory preparation and collection of dry chlorine gas
Chlorine is prepared by warming manganese(IV) oxide with concentrated hydrochloric acid:
\[\mathrm{MnO_2(s)+4HCl(aq)\rightarrow MnCl_2(aq)+Cl_2(g)+2H_2O(l)}\]
The gas is passed through saturated sodium chloride solution to remove hydrogen chloride, then through concentrated sulphuric acid to remove water vapour. Since chlorine is denser than air, it is collected by downward delivery, that is, upward displacement of air. It is not collected over water because it dissolves and reacts in water.
(a)(ii) Uses of chlorine
(b)(i) River water flowing through an industrial town may contain poisonous industrial effluents, such as acids, oils and heavy-metal salts, as well as sewage and disease-causing microorganisms. It is therefore unsafe for drinking without treatment.
(b)(ii) Uses of substances in water treatment
| Substance | Use |
|---|---|
| Sand | Filtration: removal of suspended solid particles. |
| Chlorine | Disinfection: kills germs and bacteria. |
| Calcium oxide | Neutralises acidic water and raises its pH. |
| Alum | Coagulates fine colloidal particles so that they settle or can be filtered off. |
(c) Volume of hydrogen chloride gas
\[\mathrm{2NaCl(s)+H_2SO_4(aq)\rightarrow Na_2SO_4(aq)+2HCl(g)}\]
Molar mass of \(\mathrm{NaCl}=23+35.5=58.5\ \mathrm{g\ mol^{-1}}\).
\[\text{Moles of NaCl}=\frac{5.85}{58.5}=0.100\ \mathrm{mol}\]
From the equation, \(2\) mol of \(\mathrm{NaCl}\) produce \(2\) mol of \(\mathrm{HCl}\). Therefore, moles of \(\mathrm{HCl}=0.100\ \mathrm{mol}\).
\[\text{Volume of HCl at s.t.p.}=0.100\times22.4=\boxed{2.24\ \mathrm{dm^3}}\]
(d)
(e) Differences between metals and non-metals
| Property | Metals | Non-metals |
|---|---|---|
| Physical | Good conductors of heat and electricity; malleable and ductile. | Usually poor conductors of heat and electricity; brittle when solid. |
| Chemical | Lose electrons to form positive ions and generally form basic oxides. | Gain or share electrons and generally form acidic oxides. |
Answer Details
(a)(i) Laboratory preparation and collection of dry chlorine gas
Chlorine is prepared by warming manganese(IV) oxide with concentrated hydrochloric acid:
\[\mathrm{MnO_2(s)+4HCl(aq)\rightarrow MnCl_2(aq)+Cl_2(g)+2H_2O(l)}\]
The gas is passed through saturated sodium chloride solution to remove hydrogen chloride, then through concentrated sulphuric acid to remove water vapour. Since chlorine is denser than air, it is collected by downward delivery, that is, upward displacement of air. It is not collected over water because it dissolves and reacts in water.
(a)(ii) Uses of chlorine
(b)(i) River water flowing through an industrial town may contain poisonous industrial effluents, such as acids, oils and heavy-metal salts, as well as sewage and disease-causing microorganisms. It is therefore unsafe for drinking without treatment.
(b)(ii) Uses of substances in water treatment
| Substance | Use |
|---|---|
| Sand | Filtration: removal of suspended solid particles. |
| Chlorine | Disinfection: kills germs and bacteria. |
| Calcium oxide | Neutralises acidic water and raises its pH. |
| Alum | Coagulates fine colloidal particles so that they settle or can be filtered off. |
(c) Volume of hydrogen chloride gas
\[\mathrm{2NaCl(s)+H_2SO_4(aq)\rightarrow Na_2SO_4(aq)+2HCl(g)}\]
Molar mass of \(\mathrm{NaCl}=23+35.5=58.5\ \mathrm{g\ mol^{-1}}\).
\[\text{Moles of NaCl}=\frac{5.85}{58.5}=0.100\ \mathrm{mol}\]
From the equation, \(2\) mol of \(\mathrm{NaCl}\) produce \(2\) mol of \(\mathrm{HCl}\). Therefore, moles of \(\mathrm{HCl}=0.100\ \mathrm{mol}\).
\[\text{Volume of HCl at s.t.p.}=0.100\times22.4=\boxed{2.24\ \mathrm{dm^3}}\]
(d)
(e) Differences between metals and non-metals
| Property | Metals | Non-metals |
|---|---|---|
| Physical | Good conductors of heat and electricity; malleable and ductile. | Usually poor conductors of heat and electricity; brittle when solid. |
| Chemical | Lose electrons to form positive ions and generally form basic oxides. | Gain or share electrons and generally form acidic oxides. |
Question 2 Report
(a)(i) State two differences betwecii the properties of solids and gases
(ii) What process does each of X, Y and Z represent in the changes shown below?
(b)(i) State Charles' Law (ii) Draw a sketch to graphically illustrate Charles' Law.
(c) 60 cm of hydrogen diffused through a porous membrane in 10 minutes. The same volume of a gas G diffused through the same membrane in 37.4 minutes. Determine the relative molecular mass of G. [ H = I ]
(d)(i) State two assumptions X of the kinetic theory.
(ii) Consider the reaction represented by the Solid of Liquid following equation:
H\(_{2(g)}\) + Cl\(_{2(g)}\) \(\to\) 2HCI\(_{(g)}\)
Use the kinetic theory to explain how the rate of formation of HCI\(_{(g)}\) would be affected by I. increase in temperature; II. decrease in pressure.
(e) Given different examples, mention one metal in each case vihich produces hydrogen on reacting with (i) dilute mineral acid; (ii) cold water; (iii) steam; (iv) hot, concentrated alkali.
(a)(i) Differences between solids and gases
| Solid | Gas |
|---|---|
| It has a definite shape and a definite volume. | It has no definite shape or volume and fills its container. |
| Its particles are closely packed and it is not easily compressed. | Its particles are far apart and it is highly compressible. |
(a)(ii) X is freezing (solidification); Y is sublimation; Z is condensation (liquefaction).
(b)(i) Charles' Law
At constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature.
Thus, \(V \propto T\), and \(\dfrac{V}{T}=\text{constant}\).
(b)(ii) Graphical illustration of Charles' Law
The straight line passes through the origin. For example, its gradient is \(\dfrac{160-40}{400-100}=0.40\ \text{cm}^3\text{K}^{-1}\), showing that \(V/T\) is constant.
(c) Relative molecular mass of G
By Graham's law, rate of diffusion is inversely proportional to the square root of relative molecular mass:
\[\frac{r_{H_2}}{r_G}=\sqrt{\frac{M_G}{M_{H_2}}}\]
Since equal volumes diffuse,
\[\frac{r_{H_2}}{r_G}=\frac{t_G}{t_{H_2}}=\frac{37.4}{10}=3.74\]
\[3.74=\sqrt{\frac{M_G}{2}}\]
\[M_G=2(3.74)^2=27.98\]
Therefore, the relative molecular mass of G is \(\boxed{28}\).
(d)(i) Assumptions of the kinetic theory of gases
(d)(ii) Effect on the rate of formation of HCl
\[H_{2(g)}+Cl_{2(g)}\rightarrow 2HCl_{(g)}\]
(e) Metals which produce hydrogen
Answer Details
(a)(i) Differences between solids and gases
| Solid | Gas |
|---|---|
| It has a definite shape and a definite volume. | It has no definite shape or volume and fills its container. |
| Its particles are closely packed and it is not easily compressed. | Its particles are far apart and it is highly compressible. |
(a)(ii) X is freezing (solidification); Y is sublimation; Z is condensation (liquefaction).
(b)(i) Charles' Law
At constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature.
Thus, \(V \propto T\), and \(\dfrac{V}{T}=\text{constant}\).
(b)(ii) Graphical illustration of Charles' Law
The straight line passes through the origin. For example, its gradient is \(\dfrac{160-40}{400-100}=0.40\ \text{cm}^3\text{K}^{-1}\), showing that \(V/T\) is constant.
(c) Relative molecular mass of G
By Graham's law, rate of diffusion is inversely proportional to the square root of relative molecular mass:
\[\frac{r_{H_2}}{r_G}=\sqrt{\frac{M_G}{M_{H_2}}}\]
Since equal volumes diffuse,
\[\frac{r_{H_2}}{r_G}=\frac{t_G}{t_{H_2}}=\frac{37.4}{10}=3.74\]
\[3.74=\sqrt{\frac{M_G}{2}}\]
\[M_G=2(3.74)^2=27.98\]
Therefore, the relative molecular mass of G is \(\boxed{28}\).
(d)(i) Assumptions of the kinetic theory of gases
(d)(ii) Effect on the rate of formation of HCl
\[H_{2(g)}+Cl_{2(g)}\rightarrow 2HCl_{(g)}\]
(e) Metals which produce hydrogen
Question 3 Report
(a) A solution of CuSO\(_4\) was electrolyzed between pure copper electrodes and the following results were obtained:
Mass of copper anode before experiment = 7.20 g
Mass of copper anode after experiment = 4.00 g
Mass of copper cathode before experiment = 5.75 g
From the information provided,
(i) calculate the mass of the cathode, after the experiment.
(ii) write an equation for the reaction at the I. anode, II. cathode.
(iii) state whether the colour of the solution would change during the electrolysis. Give a reason for your answer.
(iv) if the electrolysis was carried out for 1 hour 20 minutes with a current of 2.0 amperes, determine the value of the Faraday.
(b) Consider the reaction represented by the following equation:
MnO\(^-_4\) + I\(^-\) + H\(^+\) \(\to\) I\(_2\) + H\(_2\)O + Mn\(^{2+}\)
Write balanced half equation for the (i) oxidation reaction, (ii) reduction reaction.
(c)(i) Describe briefly how tin can be extracted from its ore.
(ii) State one use of tin.
(iii) Mention one property that makes tin suitable for the use stated in (c)(ii)
(d)(i) What is meant by the term pollution?
(ii) Explain why it is dangerous to run a generator in a closed room.
(a) Electrolysis of CuSO4 between pure copper electrodes
(i) Mass of cathode after the experiment
With copper electrodes the copper dissolved at the anode is deposited on the cathode, so mass lost by the anode equals mass gained by the cathode.
Mass dissolved at anode \( = 7.20 - 4.00 = 3.20\text{ g} \)
Mass of cathode after \( = 5.75 + 3.20 = \mathbf{8.95\text{ g}} \)
(ii) Electrode equations
I. Anode (oxidation): \[ Cu_{(s)} \to Cu^{2+}_{(aq)} + 2e^- \]
II. Cathode (reduction): \[ Cu^{2+}_{(aq)} + 2e^- \to Cu_{(s)} \]
(iii) The colour of the solution does not change. The blue Cu2+ ions removed at the cathode are exactly replaced by Cu2+ ions formed as the copper anode dissolves, so the concentration of the blue Cu2+ ions stays constant.
(iv) Value of the Faraday
Time \( = 1\text{ h }20\text{ min} = 80 \times 60 = 4800\text{ s} \)
Quantity of electricity \( Q = It = 2.0 \times 4800 = 9600\text{ C} \)
Moles of Cu deposited \( = \dfrac{3.20}{63.5} = 0.0504\text{ mol} \)
Each Cu2+ needs 2 electrons, so moles of electrons \( = 2 \times 0.0504 = 0.1008\text{ mol} \)
Faraday \( = \dfrac{9600}{0.1008} = \mathbf{9.52 \times 10^{4}\ C\,mol^{-1}} \) (about 95 200 C, close to the accepted 96 500 C).
(b) MnO4- + I- + H+ reaction
(i) Oxidation: \[ 2I^- \to I_2 + 2e^- \]
(ii) Reduction: \[ MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O \]
(c) Extraction of tin
(i) Tin occurs chiefly as cassiterite, SnO2. The ore is concentrated by froth flotation or gravity separation, roasted in air to burn off sulphur and arsenic impurities, then the oxide is reduced with carbon (coke) in a furnace:
\[ SnO_2 + 2C \to Sn + 2CO \]
The molten tin is run off and refined.
(ii) Use of tin: coating steel to make tin-plate for food cans (or making solder and alloys such as bronze).
(iii) Property: tin is resistant to corrosion and non-toxic, so it protects the steel and does not contaminate food.
(d) Pollution
(i) Pollution is the introduction of harmful or unwanted substances (pollutants) into the environment, causing damage to living things or the surroundings.
(ii) A generator undergoes incomplete combustion in a closed room and produces the poisonous gas carbon monoxide, CO. Carbon monoxide combines irreversibly with the haemoglobin of the blood (forming carboxyhaemoglobin), preventing the blood from carrying oxygen. This leads to suffocation and death.
Answer Details
(a) Electrolysis of CuSO4 between pure copper electrodes
(i) Mass of cathode after the experiment
With copper electrodes the copper dissolved at the anode is deposited on the cathode, so mass lost by the anode equals mass gained by the cathode.
Mass dissolved at anode \( = 7.20 - 4.00 = 3.20\text{ g} \)
Mass of cathode after \( = 5.75 + 3.20 = \mathbf{8.95\text{ g}} \)
(ii) Electrode equations
I. Anode (oxidation): \[ Cu_{(s)} \to Cu^{2+}_{(aq)} + 2e^- \]
II. Cathode (reduction): \[ Cu^{2+}_{(aq)} + 2e^- \to Cu_{(s)} \]
(iii) The colour of the solution does not change. The blue Cu2+ ions removed at the cathode are exactly replaced by Cu2+ ions formed as the copper anode dissolves, so the concentration of the blue Cu2+ ions stays constant.
(iv) Value of the Faraday
Time \( = 1\text{ h }20\text{ min} = 80 \times 60 = 4800\text{ s} \)
Quantity of electricity \( Q = It = 2.0 \times 4800 = 9600\text{ C} \)
Moles of Cu deposited \( = \dfrac{3.20}{63.5} = 0.0504\text{ mol} \)
Each Cu2+ needs 2 electrons, so moles of electrons \( = 2 \times 0.0504 = 0.1008\text{ mol} \)
Faraday \( = \dfrac{9600}{0.1008} = \mathbf{9.52 \times 10^{4}\ C\,mol^{-1}} \) (about 95 200 C, close to the accepted 96 500 C).
(b) MnO4- + I- + H+ reaction
(i) Oxidation: \[ 2I^- \to I_2 + 2e^- \]
(ii) Reduction: \[ MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O \]
(c) Extraction of tin
(i) Tin occurs chiefly as cassiterite, SnO2. The ore is concentrated by froth flotation or gravity separation, roasted in air to burn off sulphur and arsenic impurities, then the oxide is reduced with carbon (coke) in a furnace:
\[ SnO_2 + 2C \to Sn + 2CO \]
The molten tin is run off and refined.
(ii) Use of tin: coating steel to make tin-plate for food cans (or making solder and alloys such as bronze).
(iii) Property: tin is resistant to corrosion and non-toxic, so it protects the steel and does not contaminate food.
(d) Pollution
(i) Pollution is the introduction of harmful or unwanted substances (pollutants) into the environment, causing damage to living things or the surroundings.
(ii) A generator undergoes incomplete combustion in a closed room and produces the poisonous gas carbon monoxide, CO. Carbon monoxide combines irreversibly with the haemoglobin of the blood (forming carboxyhaemoglobin), preventing the blood from carrying oxygen. This leads to suffocation and death.
Question 4 Report
a) Define each of the following terms and indicate one use of each:
(i) Nuclear fission; (ii) Nuclear fusion.
(b) Alpha particle emission by \(^{293}_{25}U\) proceduces an element A. Beta particle emission by the particle A produces another element B. Element B also undergoes alpha particle emission to produce \(^{227}_{89}AC\). Write balanced equations to represent the above statement.
(c) The models below represent the filling of orbitals in an atom.
State which rule(s) is/are violated or obeyed by each model.
(d) Explain why the boiling point of H\(_2\)S with relative molecular mass of 34 is lower than that of H\(_2\)O with relative molecular mass of 18.
(e) HCI is passed into each of the following solvents:
(i) water;
(ii) methylbenzene. I. State the effect of each solution on blue litmus paper II. Compare the electrical conductivities of the two solutions.
(f) Zinc dust is added to copper (II) tetraoxosulphate (VI) solution. State;
(i) what is observed; (ii) the type of reaction that occurs.
Question 5 Report
(a) State the following laws of chemical combination: (i) Law of constant composition (ii) Law of multiple' proportion.
(b) Copper reacts with oxygen to form two oxides X and Y. On analysis, 1.535 g of X yielded 1.365 g copper and 1.450 g of Y yielded 1.160 g of cooper.
i) Determine the chemical formula of X and Y.
(ii) Calculate the mass of copper which can react with 0.500 g of oxygen to yield I. X II. Y.
(iii) Which of the laws of chemical combination is illustrated by the result in (b)(i) above. [ = 16, Cu = 63.51]
(c) Write the structure of the product responsible for the observation in each of the following reactions:
(i) A mixture of butanoic acid and ethanol warmed in the presence of concentrated H\(_2\)SO\(_4\) gives off a fragrant odour.
(ii) Sodium dissolves in propan-2-ol with effervescence to give a solution which on evaporation to dryness leaves a white precipitate.
(d) Consider the compound CH\(_3\)CH\(_2\)COOCH\(_2\)CH\(_3\).
(i) Name the compound (ii) Write the structural formula of the compound (iii) State the reagents and conditions for the formation of the compound.
(a)(i) Law of constant composition: a pure compound always contains the same elements combined in the same fixed proportion by mass.
(ii) Law of multiple proportions: when two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in a simple whole-number ratio.
(b) For X: oxygen \(= 1.535 - 1.365 = 0.170\) g. For Y: oxygen \(= 1.450 - 1.160 = 0.290\) g.
(i) X: Cu \(= \dfrac{1.365}{63.5} = 0.0215\), O \(= \dfrac{0.170}{16} = 0.0106\); ratio Cu:O \(\approx 2:1\), so X = Cu2O.
Y: Cu \(= \dfrac{1.160}{63.5} = 0.0183\), O \(= \dfrac{0.290}{16} = 0.0181\); ratio Cu:O \(\approx 1:1\), so Y = CuO.
(ii) Mass of copper combining with 0.500 g oxygen:
I. In X: \(\dfrac{1.365}{0.170} \times 0.500 = \mathbf{4.01\ g}\).
II. In Y: \(\dfrac{1.160}{0.290} \times 0.500 = \mathbf{2.00\ g}\).
(iii) The result (4.01 g and 2.00 g of copper combining with the same 0.500 g of oxygen, a 2:1 ratio) illustrates the law of multiple proportions.
(c)(i) The fragrant odour is from the ester ethyl butanoate: CH3CH2CH2COOCH2CH3.
(ii) The white solid left is sodium propan-2-oxide (sodium isopropoxide): (CH3)2CHONa, i.e. CH3CH(ONa)CH3.
(d)(i) CH3CH2COOCH2CH3 is ethyl propanoate.
(ii) Structural formula: CH3CH2COOCH2CH3.
(iii) Reagents/conditions: propanoic acid and ethanol, warmed with a little concentrated H2SO4 as catalyst (esterification, under reflux).
Answer Details
(a)(i) Law of constant composition: a pure compound always contains the same elements combined in the same fixed proportion by mass.
(ii) Law of multiple proportions: when two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in a simple whole-number ratio.
(b) For X: oxygen \(= 1.535 - 1.365 = 0.170\) g. For Y: oxygen \(= 1.450 - 1.160 = 0.290\) g.
(i) X: Cu \(= \dfrac{1.365}{63.5} = 0.0215\), O \(= \dfrac{0.170}{16} = 0.0106\); ratio Cu:O \(\approx 2:1\), so X = Cu2O.
Y: Cu \(= \dfrac{1.160}{63.5} = 0.0183\), O \(= \dfrac{0.290}{16} = 0.0181\); ratio Cu:O \(\approx 1:1\), so Y = CuO.
(ii) Mass of copper combining with 0.500 g oxygen:
I. In X: \(\dfrac{1.365}{0.170} \times 0.500 = \mathbf{4.01\ g}\).
II. In Y: \(\dfrac{1.160}{0.290} \times 0.500 = \mathbf{2.00\ g}\).
(iii) The result (4.01 g and 2.00 g of copper combining with the same 0.500 g of oxygen, a 2:1 ratio) illustrates the law of multiple proportions.
(c)(i) The fragrant odour is from the ester ethyl butanoate: CH3CH2CH2COOCH2CH3.
(ii) The white solid left is sodium propan-2-oxide (sodium isopropoxide): (CH3)2CHONa, i.e. CH3CH(ONa)CH3.
(d)(i) CH3CH2COOCH2CH3 is ethyl propanoate.
(ii) Structural formula: CH3CH2COOCH2CH3.
(iii) Reagents/conditions: propanoic acid and ethanol, warmed with a little concentrated H2SO4 as catalyst (esterification, under reflux).
Question 6 Report
(a)(i) Draw the energy profile diagram for the reaction
\(\mathrm{H_{2(g)} + I_{2(g)} \longrightarrow 2HI_{(g)}}\) \(\Delta =\) —13 kJmol\(^3\)
(ii) If the concentration of HI increases from 0 to 0.001 mol dm\(^3}\) in 50 seconds, what is the rate of the reaction?
(b) State the type of salt represented by each of the following compounds:
(i) \(\mathrm{K_4Fe(CN)_6}\) (ii) \(\mathrm{(NH_4)_2Fe(SO_4)_2 6H_2O}\) (iii) \(\mathrm{Mg(OH)NO_3}\) (iv) \(\mathrm{NaH_2PO_4}\).
(c) Explain, giving equations, the following observation: When carbon (IV) oxide is passed into lime water, it turns milky initially but turns clear with excess carbon (IV) oxide.
(d)(i) Give one use for each of the following compounds: \(\mathrm{CaCO_3}\), \(\mathrm{CaSO_4}\), \(\mathrm{NaHCO_3}\).
(ii) State a drying agent for each of the following gases: i. \(\mathrm{NH_3}\), II. HCI Ill. \(\mathrm{SO_4}\).
(iii) Write an equation to illustrate the reaction of ammonia as a reducing agent.
(e) An industrial raw material has the following composition by mass:
| Iron | = 28.1% |
| Chlorine | = 35.7% |
| Water of crystallization | = 36.2% |
Calculate the formula for the material. [ H = 1, 0 = 16, Cl = 35.5, Fe = 56 ].
(a)(i) Energy profile diagram
The reaction is exothermic, since \(\Delta H=-13\ \text{kJ mol}^{-1}\). Therefore, the products are at a lower energy level than the reactants.
(ii) Rate of reaction
\[\text{Rate}=\frac{\text{change in concentration of HI}}{\text{time taken}}=\frac{0.001-0}{50}=2.0\times10^{-5}\ \text{mol dm}^{-3}\text{s}^{-1}.\]
(b) Types of salts
(c) Action of carbon(IV) oxide on lime water
Initially, carbon(IV) oxide reacts with lime water to form insoluble calcium trioxocarbonate(IV), \(\mathrm{CaCO_3}\), which is a white precipitate. This makes the lime water milky.
\[\mathrm{Ca(OH)_2(aq)+CO_2(g)\rightarrow CaCO_3(s)+H_2O(l)}\]
On passing excess carbon(IV) oxide, the calcium trioxocarbonate(IV) dissolves to form soluble calcium hydrogentrioxocarbonate(IV). The solution therefore becomes clear.
\[\mathrm{CaCO_3(s)+CO_2(g)+H_2O(l)\rightarrow Ca(HCO_3)_2(aq)}\]
(d)(i) Uses
(d)(ii) Suitable drying agents
(d)(iii) Ammonia as a reducing agent
Ammonia reduces copper(II) oxide to copper:
\[\mathrm{2NH_3(g)+3CuO(s)\rightarrow 3Cu(s)+N_2(g)+3H_2O(l)}\]
(e) Formula of the industrial raw material
| Constituent | Mass in 100 g | Moles | Mole ratio |
|---|---|---|---|
| Fe | 28.1 g | \(28.1/56=0.502\) | \(0.502/0.502=1\) |
| Cl | 35.7 g | \(35.7/35.5=1.006\) | \(1.006/0.502=2\) |
| \(\mathrm{H_2O}\) | 36.2 g | \(36.2/18=2.011\) | \(2.011/0.502=4\) |
Hence, the mole ratio is \(\mathrm{Fe:Cl:H_2O}=1:2:4\).
\[\boxed{\mathrm{FeCl_2\cdot4H_2O}}\]
Answer Details
(a)(i) Energy profile diagram
The reaction is exothermic, since \(\Delta H=-13\ \text{kJ mol}^{-1}\). Therefore, the products are at a lower energy level than the reactants.
(ii) Rate of reaction
\[\text{Rate}=\frac{\text{change in concentration of HI}}{\text{time taken}}=\frac{0.001-0}{50}=2.0\times10^{-5}\ \text{mol dm}^{-3}\text{s}^{-1}.\]
(b) Types of salts
(c) Action of carbon(IV) oxide on lime water
Initially, carbon(IV) oxide reacts with lime water to form insoluble calcium trioxocarbonate(IV), \(\mathrm{CaCO_3}\), which is a white precipitate. This makes the lime water milky.
\[\mathrm{Ca(OH)_2(aq)+CO_2(g)\rightarrow CaCO_3(s)+H_2O(l)}\]
On passing excess carbon(IV) oxide, the calcium trioxocarbonate(IV) dissolves to form soluble calcium hydrogentrioxocarbonate(IV). The solution therefore becomes clear.
\[\mathrm{CaCO_3(s)+CO_2(g)+H_2O(l)\rightarrow Ca(HCO_3)_2(aq)}\]
(d)(i) Uses
(d)(ii) Suitable drying agents
(d)(iii) Ammonia as a reducing agent
Ammonia reduces copper(II) oxide to copper:
\[\mathrm{2NH_3(g)+3CuO(s)\rightarrow 3Cu(s)+N_2(g)+3H_2O(l)}\]
(e) Formula of the industrial raw material
| Constituent | Mass in 100 g | Moles | Mole ratio |
|---|---|---|---|
| Fe | 28.1 g | \(28.1/56=0.502\) | \(0.502/0.502=1\) |
| Cl | 35.7 g | \(35.7/35.5=1.006\) | \(1.006/0.502=2\) |
| \(\mathrm{H_2O}\) | 36.2 g | \(36.2/18=2.011\) | \(2.011/0.502=4\) |
Hence, the mole ratio is \(\mathrm{Fe:Cl:H_2O}=1:2:4\).
\[\boxed{\mathrm{FeCl_2\cdot4H_2O}}\]
Would you like to proceed with this action?