OCR A Jun 2014 Paper 1 Q3 (with explained solutions) |

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OCR A Jun 2014 Paper 1 Q3

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PMT 3 An aqueous solution of aluminium chloride can be prepared by the redox reaction between aluminium metal and dilute hydrochloric acid. A student reacts 0.0800 mol of aluminium completely with dilute hydrochloric acid to form an aqueous solution of aluminium chloride. The equation for this reaction is shown below. 2Al(s) + 6HCl(aq) 2Al Cl3(aq) + 3H2(g) (a) In terms of electron transfer, explain whether aluminium is being oxidised or reduced.[1] (b) Calculate the volume of hydrogen gas formed, in dm3, at room temperature and pressure. volume of hydrogen gas formed =dm3 [2] (c) Calculate the mass of Al Cl3 formed. Give your answer to three significant figures. mass of Al Cl3 formed =g [2] (d) Calculate the volume, in cm3, of 1.20 mol dm3 hydrochloric acid needed to react completely with 0.0800 mol of aluminium. volume =cm3 [2] [Total: 7]OCR 2014 Turn over<br />

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F321 Mark Scheme June 2014 Question (a) Answer Oxidised AND because aluminium has lost (three) electrons (cid:57) Mark Guidance ALLOW donated for lost IGNORE where electrons are transferred to IGNORE Al (cid:198) Al3+ + 3e- DO NOT ALLOW an electron or incorrect number of electrons ALLOW ECF from incorrectly calculated moles of H2 0.08 x 24 = 1.92 gets 1 mark If there is an alternative answer, check to see if there is any ECF credit possible using working below ALLOW ECF for incorrect molar mass of AlCl3 multiplied by 0.0800 and correctly rounded to 3 significant figures If there is an alternative answer, check to see if there is any ECF credit possible using working below ALLOW ECF for incorrect mol of HCl x 1000/1.20 ALLOW 66.7 (66.67 or 66.667 etc) for 1 mark DO NOT ALLOW 66.6 (66.66 or 66.666 etc) (b) (c) (d) FIRST CHECK THE ANSWER ON THE ANSWER LINE IF answer = 2.88 dm3 award 2 marks Mol of H2 = 0.12 (cid:57) Volume of H2 = 0.12 x 24.0 = 2.88 dm3 (cid:57) FIRST CHECK THE ANSWER ON THE ANSWER LINE IF answer = 10.7 g award 2 marks Correctly calculates molar mass of AlCl3 = 133.5 g (cid:57) Mass of AlCl3 formed = 0.0800 x 133.5 = 10.7 (g) (cid:57) FIRST CHECK THE ANSWER ON THE ANSWER LINE IF answer = 200(.0) cm3 award 2 marks Correctly calculates moles of HCl needed = 0.0800 x 3 = 0.24(0) mol (cid:57) Volume of HCl = 0.24(0) x 1000/1.2 = 200 cm3 (cid:57) Total<br />

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