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Pergunta 1 Relatório
Consider the following experimental set-up.
(i) Identify by name: P; Q; R; S and T.
(ii) State the method of collection of gas, S.
(iii) What is the function of R in the experimental set-up?
(iv) Write the balanced equation of the reaction for the preparation of gas S.
(i) Identification of the labelled parts:
(ii) Method of collection of gas S: Ammonia is collected by upward delivery (downward displacement of air), because it is less dense than air and is very soluble in water.
(iii) Function of R (calcium oxide): It dries the gas, removing water vapour from the ammonia before collection. Calcium oxide is used because it does not react with ammonia (an acidic drying agent such as concentrated H2SO4 or anhydrous CaCl2 would react with it).
(iv) Balanced equation for the preparation of gas S:
Ca(OH)2(aq) + 2NH4Cl(aq) → CaCl2(aq) + 2H2O(l) + 2NH3(g)
Detalhes da Resposta
(i) Identification of the labelled parts:
(ii) Method of collection of gas S: Ammonia is collected by upward delivery (downward displacement of air), because it is less dense than air and is very soluble in water.
(iii) Function of R (calcium oxide): It dries the gas, removing water vapour from the ammonia before collection. Calcium oxide is used because it does not react with ammonia (an acidic drying agent such as concentrated H2SO4 or anhydrous CaCl2 would react with it).
(iv) Balanced equation for the preparation of gas S:
Ca(OH)2(aq) + 2NH4Cl(aq) → CaCl2(aq) + 2H2O(l) + 2NH3(g)
Pergunta 2 Relatório
Burette readings (initial and final) must be given to two decimal places. Volume of pipette used must also be recorded but no account of experimental procedure is required. All calculations must be done in your answer book.
A is a solution of hydrochloric acid. B is a solution containing 2.45g of anhydrous sodium trioxocarbonate (IV) in 250g of solution.
(a) Put A into the burette and titrate it against 20.0 cm\(^3\) or 25.0 cm\(^3\) portion of B using methyl orange as an indicator. Repeat the exercise to obtain consistent titres. Tabulate your burette reading and calculate the average volume of A used. The equation for the reaction involved in the titration is Na\(_2\)CO\(_{3(aq)}\) + 2HCI\(_{(aq)}\) + H\(_2\)O\(_{(l)}\)
(b) From your results and the information provided, calculate the:
(i) concentration of B in moldm\(^{-3}\)
(ii) concentration of A in moldm\(^{-3}\)
(iii) concentration of A in gdm\(^{-3}\)
(iv) volume of the gas evolved in the reaction at s.t.p.
[H = 1.00; C: 12.0; O = 16.0; Na = 23.0; Na = 23.0; Cl = 35.5; Molar Volume = 22.4 dm\(^3\)mol\(^{-3}\)]
(a) Titration
A (hydrochloric acid) is in the burette; 25.0 cm3 portions of B are pipetted into the flask with methyl orange. A specimen concordant set of readings:
| Burette readings (cm3) | Rough | 1st | 2nd |
|---|---|---|---|
| Final reading | 23.40 | 23.15 | 23.10 |
| Initial reading | 0.00 | 0.00 | 0.00 |
| Volume of A used | 23.40 | 23.15 | 23.10 |
Average volume of A \(=\dfrac{23.15+23.10}{2}=23.13\ \text{cm}^3\)
(b)(i) Concentration of B in mol dm-3 (M(Na2CO3) = 106)
B contains 2.45 g in 250 cm3, i.e. \(2.45\times4=9.80\) g dm-3.
\[ [B]=\frac{9.80}{106}=0.0925\ \text{mol dm}^{-3} \](ii) Concentration of A in mol dm-3 (equation: Na2CO3 + 2HCl → 2NaCl + H2O + CO2)
\[ n(B)=\frac{0.0925\times25.0}{1000}=2.313\times10^{-3}\ \text{mol} \] \[ n(\text{HCl})=2\times2.313\times10^{-3}=4.625\times10^{-3}\ \text{mol} \] \[ [A]=\frac{4.625\times10^{-3}\times1000}{23.13}=0.200\ \text{mol dm}^{-3} \](iii) Concentration of A in g dm-3 (M(HCl) = 36.5)
\[ 0.200\times36.5=7.30\ \text{g dm}^{-3} \](iv) Volume of CO2 at s.t.p. (from the 25.0 cm3 of B used)
\[ n(\text{CO}_2)=n(\text{Na}_2\text{CO}_3)=2.313\times10^{-3}\ \text{mol} \] \[ V=2.313\times10^{-3}\times22400=51.8\ \text{cm}^3\ (0.0518\ \text{dm}^3) \]The titre figures are specimen readings; substitute your own concordant burette values.
Detalhes da Resposta
(a) Titration
A (hydrochloric acid) is in the burette; 25.0 cm3 portions of B are pipetted into the flask with methyl orange. A specimen concordant set of readings:
| Burette readings (cm3) | Rough | 1st | 2nd |
|---|---|---|---|
| Final reading | 23.40 | 23.15 | 23.10 |
| Initial reading | 0.00 | 0.00 | 0.00 |
| Volume of A used | 23.40 | 23.15 | 23.10 |
Average volume of A \(=\dfrac{23.15+23.10}{2}=23.13\ \text{cm}^3\)
(b)(i) Concentration of B in mol dm-3 (M(Na2CO3) = 106)
B contains 2.45 g in 250 cm3, i.e. \(2.45\times4=9.80\) g dm-3.
\[ [B]=\frac{9.80}{106}=0.0925\ \text{mol dm}^{-3} \](ii) Concentration of A in mol dm-3 (equation: Na2CO3 + 2HCl → 2NaCl + H2O + CO2)
\[ n(B)=\frac{0.0925\times25.0}{1000}=2.313\times10^{-3}\ \text{mol} \] \[ n(\text{HCl})=2\times2.313\times10^{-3}=4.625\times10^{-3}\ \text{mol} \] \[ [A]=\frac{4.625\times10^{-3}\times1000}{23.13}=0.200\ \text{mol dm}^{-3} \](iii) Concentration of A in g dm-3 (M(HCl) = 36.5)
\[ 0.200\times36.5=7.30\ \text{g dm}^{-3} \](iv) Volume of CO2 at s.t.p. (from the 25.0 cm3 of B used)
\[ n(\text{CO}_2)=n(\text{Na}_2\text{CO}_3)=2.313\times10^{-3}\ \text{mol} \] \[ V=2.313\times10^{-3}\times22400=51.8\ \text{cm}^3\ (0.0518\ \text{dm}^3) \]The titre figures are specimen readings; substitute your own concordant burette values.
Pergunta 3 Relatório
Credit will be given for strict adherence to the instructions, for observations, precisely recorded, and for accurate inferences. All tests, observations, and inferences must be clearly entered in your answer book, in ink, at the time they are made.
C contains two cations and two anions. Perform the following exercises on C. Record your observations and identify any gas (es) evolved. State the conclusion you draw from the result of each test.
(a) Dissolve all of C in about 10 cm\(^3\) of distilled water. Stir the resulting solution thoroughly.
(i) To about 2 cm\(^3\) of the solution, add few drops of AgNO\(_3\) solution, followed by HNO\(_{3(aq)}\). To the mixture, add excess NH\(_{3(aq)}\)
(ii) To another 2 cm\(^3\) portion of the solution, add dil. HCl followed by BaCl\(_2\) solution.
(iii) To another 2 cm\(^3\) portion of the solution, add NaOH\(_{3(aq)}\) dropwise and then in excess. Warm the mixture.
(iv) To another 2 cm\(^3\) portion of the solution, add NH\(_{3(aq)}\) dropwise and then in excess.
Result for C
All of C dissolves in about 10 cm3 of distilled water to give a clear solution.
| Test | Observation | Inference / conclusion |
|---|---|---|
| (i) To 2 cm3 of C(aq), add AgNO3(aq), followed by HNO3(aq), then excess NH3(aq). | A white precipitate forms with AgNO3(aq). The precipitate is insoluble in dilute HNO3(aq), but dissolves in excess NH3(aq). No gas is evolved. | Chloride ion, Cl−, is present. The precipitate is AgCl. |
| (ii) To another 2 cm3 portion, add dilute HCl(aq), followed by BaCl2(aq). | No visible reaction and no gas is evolved on adding dilute HCl. A white chalky precipitate forms on adding BaCl2(aq). | Sulphate ion, SO42−, is present. The precipitate is BaSO4. |
| (iii) To another 2 cm3 portion, add NaOH(aq) dropwise, then in excess. Warm the mixture. | A white gelatinous precipitate forms on adding NaOH(aq) dropwise. It dissolves in excess NaOH(aq). On warming, a colourless pungent gas is evolved. The gas turns damp red litmus paper blue and forms dense white fumes with hydrogen chloride gas. | Zn2+ or Al3+ is initially suspected from the amphoteric white precipitate. The gas is ammonia, NH3; hence NH4+ is present. |
| (iv) To another 2 cm3 portion, add NH3(aq) dropwise, then in excess. | A white gelatinous precipitate forms on adding NH3(aq) dropwise. The precipitate dissolves in excess NH3(aq). | Zn2+ is present. Zinc hydroxide dissolves in excess ammonia to form a colourless complex. |
Final conclusion: C contains the cations Zn2+ and NH4+, and the anions Cl− and SO42−.
Detalhes da Resposta
Result for C
All of C dissolves in about 10 cm3 of distilled water to give a clear solution.
| Test | Observation | Inference / conclusion |
|---|---|---|
| (i) To 2 cm3 of C(aq), add AgNO3(aq), followed by HNO3(aq), then excess NH3(aq). | A white precipitate forms with AgNO3(aq). The precipitate is insoluble in dilute HNO3(aq), but dissolves in excess NH3(aq). No gas is evolved. | Chloride ion, Cl−, is present. The precipitate is AgCl. |
| (ii) To another 2 cm3 portion, add dilute HCl(aq), followed by BaCl2(aq). | No visible reaction and no gas is evolved on adding dilute HCl. A white chalky precipitate forms on adding BaCl2(aq). | Sulphate ion, SO42−, is present. The precipitate is BaSO4. |
| (iii) To another 2 cm3 portion, add NaOH(aq) dropwise, then in excess. Warm the mixture. | A white gelatinous precipitate forms on adding NaOH(aq) dropwise. It dissolves in excess NaOH(aq). On warming, a colourless pungent gas is evolved. The gas turns damp red litmus paper blue and forms dense white fumes with hydrogen chloride gas. | Zn2+ or Al3+ is initially suspected from the amphoteric white precipitate. The gas is ammonia, NH3; hence NH4+ is present. |
| (iv) To another 2 cm3 portion, add NH3(aq) dropwise, then in excess. | A white gelatinous precipitate forms on adding NH3(aq) dropwise. The precipitate dissolves in excess NH3(aq). | Zn2+ is present. Zinc hydroxide dissolves in excess ammonia to form a colourless complex. |
Final conclusion: C contains the cations Zn2+ and NH4+, and the anions Cl− and SO42−.
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