Edexcel Jun 2014 (R) Paper 1 Q20 (with explained solutions) |

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Edexcel Jun 2014 (R) Paper 1 Q20

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20 In an experiment to make crystals of hydrated copper(II) nitrate, a sample of 5.60 g of copper(II) oxide was added to 50 cm3 of 2.50 mol dm3 nitric acid. The following reaction occurred. CuO(s) + 2HNO3(aq) (cid:111) Cu(NO3)2(aq) + H2O(l) (a) Calculate the number of moles of each reactant present, and use this to show that the copper(II) oxide was in excess. The molar mass of copper(II) oxide, CuO, is 79.5 g mol1. (3) Moles of copper(II) oxide added Moles of nitric acid used The copper(II) oxide is in excess because*P42990A01528* 15 Turn over<br />
(b) The copper(II) nitrate solution was heated gently to concentrate it, and then left to crystallize. The mass of hydrated copper(II) nitrate crystals, Cu(NO3)2.6H2O, obtained was 12.52 g. Calculate the percentage yield. The molar mass of Cu(NO3)2.6H2O is 295.6 g mol1. (c) Give one reason why the percentage yield is less than 100%, even though the nitric acid was completely reacted. (3) (1)*(d) (i) The nitrate ion, NO3 , contains both covalent and dative covalent bonds. What is the difference between these types of bond? (2)16 *P42990A01628*<br />
(ii) Complete the dot and cross diagram to show the bonding in the nitrate ion. Only the outer electron shells for each atom need to be shown. Represent the nitrogen electrons with crosses (), and oxygen electrons with dots, ((cid:120)). The symbol * on the diagram represents the extra electron giving the ion its charge. (3)*O(Total for Question 20 = 12 marks) *P42990A01728* 17 Turn over<br />
BLANK PAGE 18 *P42990A01828*<br />

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Question Number 20(a) Acceptable Answers Reject 1st Mark Mol CuO = (5.60/79.5) = 0.07044 / 0.0704 / 0.070 / 0.07 (1) 2nd Mark Mol of nitric acid = (50 x 2.50/1000) = 0.125 (1) 3rd Mark Reacting ratio =2:1 and nitric acid less than double moles of copper oxide/ Reacting ratio =2:1 and copper oxide more than half of moles of nitric acid OR moles acid needed to react with all CuO = (2 x 0.070 =) 0.140 which is more than 0.125 OR 0.125 mol nitric acid can only react with 0.0625 mol CuO (1) Mark 6CH0101R 1406<br />
Mark Acceptable Answers Reject 1st Mark Moles product = 0.5 x 0.125 =0.0625 (1) Allow TE from moles HNO3 2nd Mark Theoretical yield = (0.0625 x 295.6 = ) 18.475 g (1) Allow ECF on multiplying moles product by 295.6 3rd Mark % yield = (12.52/18.475 x 100) = 67.767 / 67.8 / 68 (1) Alternative route for 2nd and 3rd Marks mol product = (12.52 / 295.6) = 0.04235 (1) % yield = (0.04235/0.0625 x 100 = 67.767 / 67.8/ 68 (1) TE from (a) If moles of product taken as 0.125, final answer = 33.88% which scores (2) TE for calculation based on moles of copper(II) oxide which gives an answer between 60.128% and 60.506% max(2) 4.24% scores (0) overall Question Number 20(b) 6CH0101R 1406<br />
Question Number 20(c) Mark Acceptable Answers Reject Some product remains in solution/ some product does not crystallize Allow loss of material on transferring, if explained, such as Crystals remain in / on filter paper Spitting (of solution on heating) IGNORE References to impure reactants Incomplete reaction Just experimental error solution evaporates Acceptable Answers Question Number 20(d)(i) Covalent bond: (shared pair of electrons Reject Mark using) one electron from each atom (1) Dative covalent bond: (shared pair of electrons using) two electrons from same atom (1) Question Number 20(d)(ii) Double bond between N and one oxygen atom (1) Acceptable Answers Single bond between N and O* (1) Dative single bond between N and one O atom (1) Reject Mark Max 2 if any lone pair electrons are missing from any of the three oxygen atoms. Total for Q20 = 12 marks 6CH0101R 1406<br />

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