OCR A Jun 2014 Paper 5 Q8 (with explained solutions) |

Skip to content

OCR A Jun 2014 Paper 5 Q8

Answers available below

8 A student carries out an investigation to prepare and analyse a sample of barium ferrate(VI), BaFeO4. The steps in the investigation are shown below. 18 Step 1 The student adds solid iron(III) oxide to a hot aqueous solution containing an excess of hydroxide ions. The student bubbles chlorine gas through the mixture. A solution forms containing aqueous ferrate(VI) ions, FeO4 2(aq), and aqueous chloride ions. Step 2 The student adds aqueous barium chloride to the resulting solution. A precipitate of impure barium ferrate(VI) forms. The precipitate is filtered, washed with distilled water and dried. The student obtains 0.437 g of impure solid barium ferrate(VI). Step 3 An excess of acidified aqueous potassium iodide is added to the solid from step 2. The BaFeO4 reacts as shown below, and the impurity does not react. A solution forms containing aqueous iodine, I2(aq). BaFeO4(s) + 8H+(aq) + 3I(aq) Fe3+(aq) + Ba2+(aq) + 1I2(aq) + 4H2O(l) Step 4 The student determines the amount of I2 formed by carrying out a titration with aqueous sodium thiosulfate, Na2S2O3(aq). 2S2O3 2(aq) + I2(aq) S4O6 2(aq) + 2I(aq) 26.4 cm3 of 0.100 mol dm3 Na2S2O3(aq) are required to reach the end point. (a) Construct an equation for the oxidation of iron(III) oxide (step 1).[2] (b) Write an ionic equation for the formation of barium ferrate(VI) (step 2). Include state symbols.[1]OCR 2014<br />
(c) In step 3, what is the reducing agent? Explain your answer in terms of electrons. 19 reducing agentexplanation[2] (d) The solid sample of barium ferrate(VI) obtained in step 2 is impure. Determine the percentage, by mass, of barium ferrate(VI) in the 0.437 g of solid formed in step 2. Give your answer to one decimal place. percentage of barium ferrate(VI) =% [4] Question 8 continues on page 20OCR 2014 Turn over<br />
(e) When the solution is not alkaline, ferrate(VI) ions react with water. The reaction forms a gas with a density of 1.333103 g cm3, measured at room temperature and pressure, and an orangebrown precipitate. 20Determine the formulae of the gas and the precipitate.Write an equation for the reaction that takes place. gasprecipitateequation[3] [Total: 12] END OF QUESTION PAPEROCR 2014<br />

Show answer

F325 Question (a) (b) (c) Mark Scheme June 2014 Marks Guidance 2 + 5H2O + 6e 2 + 6Cl + 5H+ 2 + 5HCl + Cl ALLOW multiples ALLOW oxidation half equation for two marks Fe2O3 + 10OH2FeO4 Correct species would obtain 1 markquestion: equation for oxidation ALLOW variants forming H+ for 1 mark, e.g: Fe2O3 + 3Cl2 + 5OH2FeO4 Fe2O3 + 3Cl2 + 5OH2FeO4 Balanced ionic equation AND state symbols required DO NOT ALLOW +2 or 2 for ionic charges IGNORE H+ OR acidified ALLOW iodide/potassium iodide but DO NOT ALLOW iodine ALLOW I loses electrons AND to form I2 ALLOW Fe(6+) OR Fe6+ Answer Fe2O3 + 3Cl2 + 10OH2FeO4 First mark for all 6 species Second mark for balancing 2 + 6Cl + 5H2O (cid:57)(cid:57) 2(aq)BaFeO4(s) (cid:57) Ba2+(aq) + FeO4 Reason can ONLY be correct from correct reducing agent --------------------------------------------------------------------------- reducing agent: I OR KI (cid:57) I adds/donates/loses electrons AND to FeO4 ALLOW Fe(6+) OR Fe6+ 2 OR to BaFeO4 OR to Fe(VI) or to Fe(+6) (cid:57) 27<br />
F325 (d) Mark Scheme June 2014 2) =x 2/3 x 2.64 (cid:117) 103 = 8.8(0) (cid:117) 104 (mol) (cid:57) = 2.64 (cid:117) 103 (mol) (cid:57) 2) used = 0.1000 (cid:117) 26.4 1000 FIRST, CHECK THE ANSWER ON ANSWER LINE IF answer = 51.8%, award 4 marks. -------------------------------------------------------------------- n(S2O3 n(FeO4 Mass BaFeO4 in sample = 8.8 (cid:117) 104 (cid:117) 257.1 g = 0.226248 g (cid:57) % purity = 0.226248 0.437 MUST be to one decimal place (in the question) ------------------------------------------------------------ As an alternative for the final two marks, ALLOW: Theoretical amount of BaFeO4 = 0.437 257.1 % purity = 8.8 (cid:117) 10(cid:16)4 1.70 (cid:117) 10(cid:16)3 (cid:117)100 = 51.8% (cid:57) (cid:117)100 = 51.8% (cid:57) = 0.00170 (mol) (cid:57) 28 FULL ANNOTATIONS MUST BE USED -------------------------------------------------------------------------------- For alternative answers, look first at common ECFs below. Then check for ECF credit possible using working below IF a step is omitted but subsequent step subsumes previous, then award mark for any missed step ----------------------------------------------------------- Working must be to at least 3 SF throughout until final % mark BUT ignore trailing zeroes, ie for 0.880 allow 0.88 ECF answer above (cid:117)(cid:117) 2/3 This mark may be seen in 2 steps via I2 but the mark is for both steps combined ECF 257.1 (cid:117) answer above ECF answer above ALLOW 51.7% FROM 0.226 g BaFeO4 (earlier rounding) ----------------------------------------------------------------------------------- Common ECFs: No (cid:117) 2/3 for n(FeO4 No2 for n(FeO4 24.6 used instead of 26.4: % purity = 25.9% % purity = 77.7%/77.6% 3 marks 3 marks % purity = 48.2% (cid:117)100 0.437 2): 2): 3 marks<br />
F325 (e) Mark Scheme June 2014 3 12 DO NOT ALLOW names IGNORE a balancing number shown before a formula ALLOW Fe(OH)3(H2O)3 ALLOW multiples ALLOW 2FeO4 2 + 11H2O1O2 + 2Fe(OH)3(H2O)3 + 4OH gas: O2 (cid:57) precipitate: Fe(OH)3 (cid:57) equation: 2FeO4 OR 2FeO4 2 + 5H2O1O2 + 2Fe(OH)3 + 4OH 2 + H2O + 4H+1O2 + 2Fe(OH)3 (cid:57) Total 29<br />

Question 1 2 3 4 5 6 7 8

About us | Need Help? | Feedback

Proudly powered by WordPress | Theme: Sydney by aThemes.