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CIE Jun 2013 v2 Paper 2 Q2
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![Examiners For Use Chile saltpetre is a mineral found in Chile and Peru, and which mainly consists of sodium nitrate, NaNO3. The mineral is puri ed to concentrate the NaNO3 which is used as a fertiliser and in somereworks. In order tond the purity of a sample of sodium nitrate, the compound is heated in NaOH(aq) with Devarda's alloy which contains aluminium. This reduces the sodium nitrate to ammonia which is boiled off and then dissolved in acid. 3NaNO3(aq) + 8Al (s) + 5NaOH(aq) + 18H2O(l)3NH3(g) + 8NaAl (OH)4(aq) The ammonia gas produced is dissolved in an excess of H2SO4 of known concentration. 2NH3 + H2SO4(NH4)2SO4 The amount of unreacted H2SO4 is then determined by back-titration with NaOH of known concentration. H2SO4 + 2NaOHNa2SO4 + 2H2O (a) A 1.64 g sample of impure NaNO3 was reacted with an excess of Devarda's alloy. The NH3 produced was dissolved in 25.0 cm3 of 1.00 mol dm3 H2SO4. When all of the NH3 had dissolved, the resulting solution was titrated with NaOH(aq). For neutralisation, 16.2 cm3 of 2.00 mol dm3 NaOH were required. (i) Calculate the amount, in moles, of H2SO4 present in the 25.0 cm3 of 1.00 mol dm3 H2SO4. (ii) Calculate the amount, in moles, of NaOH present in 16.2 cm3 of 2.00 mol dm3 NaOH. (iii) Use your answer to (ii) to calculate the amount, in moles, of H2SO4 that reacted with 16.2 cm3 of 2.00 mol dm3 NaOH. (iv) Use your answers to (i) and (iii) to calculate the amount, in moles, of H2SO4 that reacted with the NH3.UCLES 2013 9701/22/M/J/13<br />
Examiners For Use (v) Use your answer to (iv) to calculate the amount, in moles, of NH3 that reacted with the H2SO4. (vi) Use your answer to (v) to calculate the amount, in moles, of NaNO3 that reacted with the Devarda's alloy. (vii) Hence calculate the mass of NaNO3 that reacted. (viii) Use your answer to (vii) to calculate the percentage by mass of NaNO3 present in the impure sample. Write your answer to a suitable number of signi cantgures. (b) The above reaction is an example of a redox reaction. What are the oxidation numbers of nitrogen in NaNO3 and in NH3? NaNO3NH3[9] [1] [Total: 10]UCLES 2013 9701/22/M/J/13 [Turn over<br />](https://learnah.org/wp-content/uploads/2018/08/June_2013_v2_QP_-_Paper_2_CIE_Chemistry_A-level_2-0.jpg)
![Mark Scheme GCE AS/A LEVELMay/June 2013 Syllabus 9701 Paper 22 Page 3 (a) (i) n(H2SO4) = (ii) n(NaOH) = 001025 1000 002216 1000 025 mol 0324 mol (1) (1) (1) (1) (1) (1) (1) (1) (1) [9] (1) [1] [Total: 10] (iii) n(H2SO4) reacting with NaOH = 0324 0162 mol (iv) n(H2SO4) reacting with NH3 = 0.025 - 0.0162 = 0.0088 mol (v) n(NH3) reacting with H2SO4 = 2 x 0.0088 = 0.0176 mol (vi) n(NaNO3) reacting = n(NH3) produced = 0.0176 mol (vii) mass of NaNO3 that reacted = 0.0176 x 85 = 1.496 g (viii) % of NaNO3 = 496 100 641 91 2195122 291 give one mark for the correct expression give one mark for answer given as 91.2i.e to 3 sig. fig. allow ecf where appropriate (b) NaNO3 +5 and NH3 -3 both requiredCambridge International Examinations 2013<br />](https://learnah.org/wp-content/uploads/2018/08/June_2013_v2_MS_-_Paper_2_CIE_Chemistry_A-level_2.jpg)