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Class practice 38.4

Propanone reacts with iodine in the presence of hydrochloric acid. CH 3 COCH 3 (aq) + I 2 (aq)  CH 3 COCH 2 I(aq) + H + (aq) + I – (aq). H + (aq). Class practice 38.4.

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Class practice 38.4

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  1. Propanone reacts with iodine in the presence of hydrochloric acid. CH3COCH3(aq) + I2(aq)  CH3COCH2I(aq) + H+(aq) + I–(aq) H+(aq) Class practice 38.4

  2. A reaction mixture of propanone, iodine and dilute hydrochloric acid was prepared. At regular time intervals, 10.0 cm3 of the reaction mixture was withdrawn and poured into excess sodium hydrogencarbonate solution. The third portion required 6.5 cm3 of 0.005 M sodium thiosulphate solution, Na2S2O3(aq), to react completely with the unreacted iodine.(a) Suggest a piece of glass apparatus used to deliver the small portions of the reaction mixture. (a)Pipette (10.0 cm3 type)

  3. (b) Explain the purpose of adding the small portions of the reaction mixture to excess sodium hydrogencarbonate solution. Write an equation for the reaction involved in this step. (b)To quench the reaction by removing the acid catalyst: NaHCO3(aq) + HCl(aq)  NaCl(aq) + CO2(g) + H2O(l)

  4. (ii)No. of moles of Na2S2O3(aq) used 6.5 1000 = 0.005  mol (c) (i) Balance the following ionic equation for the titration reaction. S2O32–(aq) + I2(aq)  S4O62–(aq) + I–(aq) (ii) Calculate the concentration of iodine in the third portion. (c)(i)2S2O32(aq) + I2(aq)  S4O62(aq) + 2I(aq)

  5. = 3.25  10−5 mol From the equation in (i), mole ratio of S2O32 :I2= 2: 1.  no. of moles of I2 present 3.25  10−5 2 = 1.625  10−5 mol Concentration of I2 in the third portion 1.625  10−5 10.0  10−3 = 1.625  10−3 mol dm−3 = mol = mol dm−3

  6. (d) Sketch a graph showing the concentration of iodine (in mol dm–3) against time (in minute).

  7. (d) Concentration of I2 (mol dm−3) 0 Time (min)

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