DHS_H1_CHEM_P2_Answer_Scheme

© DHS 2015 8872/02 [Turn over Section A Answer all questions in the spaces provided. 1 (a) Calcium ethanedioate, CaC 2O4, is a white needle–like crystalline solid. When a pure sample of anhydrous CaC 2O4 was heated strongly at 400 C until no further change in mass is observed, a white solid B and 0.028 g of carbon monoxide gas were obtained as the only products. (i) Given that 1 mole of CaC 2O4 decomposes to give 1 mole of CO(g), write a balanced equation for the thermal decomposition of CaC2O4 and identify solid B. equation : CaC2O4 CaCO3 + CO Solid B : CaCO3 (ii) Hence, determine the mass of B obtained in the reaction. Moles of of CaCO3 = moles of CO = 028 0280 . . = 0.0010 mol Mass of CaCO3 obtained = 0.0010  100.0 = 0.10 g [3] (b) The formula of potassium hydrogen ethanedioate can be written as K xHy(C2O4)z. In an experiment to determine the values of x, y and z, 4.50 g of this compound was dissolved in water and the solution made up to 1 dm3. 20.0 cm 3 of the solution was pipetted into a conical flask and then titrated with 0.0200 mol dm ––3 KMnO4 in an acidic medium. It was found that 16.50 cm 3 of KMnO 4(aq) was needed for the complete reaction with ethanedioate ions, C 2O4 2–, present. During the titration, effervescence of carbon dioxide is produced. Given the reaction of manganate(VII) in acidic medium as: MnO4 – + 8H+ + 5e– Mn2+ +4H2O (i) Write a balanced half–equation for the reaction of C2O4 2 during titration. half–equation : C2O4 2 2CO2 + 2e– (ii) Calculate the mass of C2O4 2– present in 1 dm3 of the solution. E.c.f. if half–equation from (i) is wrong. Overall equation: 5C2O4 2 + 2MnO4  + 16H+ 10CO2 + 2Mn2+ + 8H2O Moles of C2O4 2 in 20.0 cm3 = 2 5  (0.0200  1000 50.16 ) = 8.25  10–4 mol

© DHS 2015 8872/02 [Turn over Mass of C2O4 2 in 1000 cm3 = (8.25  104 )  88.0  20 1000 = 3.63 g (iii) Given that 4.50 g of potassium hydrogen ethanedioate contains 0.060 g of hydrogen, calculate the mass of potassium present in the sample. Mass of potassium present = 4.50 – 0.060 –3.63 = 0.81 g (iv) Hence, determine the values x, y and z. K H C2O4 Mass / g 0.81 0060 3.63 Moles / mol 0.81/ 39.1 = 0.021 0.060/1.0 = 0.060 3.63/88.0 = 0.041 Ratio 1 3 2 Working to calculate moles of all species: x = 1, y = 3 and z = 2 Formula is KH3(C2O4)2. [6] [Total: 9]

© DHS 2015 8872/02 [Turn over 2 The following table lists the standard enthalpy changes of combustion, Hc , of some monohydric alcohols. alcohol Hc  / kJ mol–1 methanol(l) –715 ethanol(l) –1367 propan–1–ol(l) ? butan–1–ol(l) –2671 . (a) Write an equation to represent the standard enthalpy change of combustion of p