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Chemistry 1010 Loader (Fall 2004) Page 1 of 2 Chemistry 1010 Handout 4: Reactions, equations and Redox 1. Write molecular and net-ionic (where applicable) equations for the reactions that take place in the following experiments. (a) Sodium hydroxide solution reacts with zinc nitrate solution to give a precipitate of zinc hydroxide. NaOH(aq) + Zn(NO3)2(aq) → Zn(OH)2(s) + NaNO3 (aq) 2 OH–(aq) + Zn2+(aq) → Zn(OH)2(s) (b) Copper(II) oxide dissolves in dilute hydrochloric acid. CuO(s) + 2 HCl(aq) → CuCl2(aq) + H2O(l) CuO(s) + 2 H+(aq) → Cu2+(aq) + H2O(l) (c) Potassium chromate reacts with lead(II) nitrate solution to give a yellow precipitate. K2CrO4(aq) + Pb(NO3)2(aq) → PbCrO4(s) + 2 KNO3 (aq) CrO42–(aq) + Pb2+(aq) → PbCrO4(s) (d) Magnesium ribbon reacts with dilute hydrochloric acid. Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) Mg(s) + 2 H+(aq) → Mg2+(aq) + H2(g) (e) Solid calcium carbonate reacts with dilute nitric acid to give carbon dioxide gas. CaCO3(s) + 2 HNO3(aq) → Ca(NO3)2(aq) + CO2(g) + H2O(l) CaCO3(s) + 2 H+(aq) → Ca2+(aq) + CO2(g) + H2O(l) (f) Potassium metal reacts with cold water to give hydrogen gas. 2 K(s) + 2 H2O(l) → 2 KOH(aq) + H2(g) 2 K(s) + 2 H2O(l) → 2 K+(aq)+ 2 OH–(aq) + H2(g) (g) Sodium hydrogen carbonate is heated strongly in a crucible leaving a residue of sodium carbonate. 2 NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g) (h) Potassium chlorate is heated strongly with a catalyst to produce oxygen gas. 2 KClO3(s) → 2 KCl(s) + 3 O2(g) (i) P4O10(s) react and dissolves in water to give a solution that turns blue litmus paper red. P4O10(s) + 6 H2O(l) → 4 H3PO4(aq) (j) Strontium oxide dissolves in water to give a solution that turns red litmus paper blue. (k) Aqueous barium nitrate reacts with aqueous lithium sulfate to give a precipitate of solid barium sulfate and aqueous lithium nitrate. Ba(NO3)2(aq) + Li2SO4(aq) → BaSO4(s) + 2 LiNO3(aq) Ba2+(aq) + SO42–(aq) → BaSO4(s) (l) lead(II) nitrate solution is mixed with sodium sulfate solution Pb(NO3)2(aq) + Na2SO4(aq) → PbSO4(s) + 2 NaNO3(aq) SO42–(aq) + Pb2+(aq) → PbSO4(s) (m) sodium hydroxide solution is added to a solution of ammonium chloride (ammonia gas is evolved) NaOH(aq) + NH4Cl(aq) → NaCl(aq) + NH3(g) + H2O(l) OH–(aq) + NH4+(aq) → NH3(g) + H2O(l) (n) calcium reacts with water Ca(s) + 2 H2O(l) → Ca(OH)2(aq) + H2(g) Ca(s) + 2 H2O(l) → Ca2+(aq) + 2 OH–(aq) + H2(g) (o) copper(II) sulfide(s) reacts with dilute sulfuric acid CuS(s) + H2SO4(aq) → CuSO4(s) + H2S(g) 2 H+(aq) + S2–(aq) → H2S(g) (p) aluminum sulfate solution is mixed with barium hydroxide solution Al2(SO4)3(aq) + 3 Ba(OH)2(aq) → 3 BaSO4(s) + 2 Al(OH)3(s) 2 Al3+(aq) + 3 SO42–+ 3 Ba2+ + 6 OH–(aq) → 3 BaSO4(s) + 2 Al(OH)3(s) (q) lithium oxide(s) is added to water Li2O(s) + H2O(l) → 2 LiOH(aq) Li2O(s) + H2O(l) → 2 Li +(aq) + 2 OH–(aq) (r) sodium bromide solution is added to an acidified (H2SO4) potassium chromate solution 8 H2SO4(aq) + 6 NaBr(aq) + 2 K2CrO4(aq) → 3 Br2(aq) + Cr2(SO4)3(aq) + 8 H2O(l) + 2 K2SO4(aq) + 3 Na2SO4(aq) 16 H+(aq) + 6 Br–(aq) + 2 CrO42–(aq) → 3 Br2(aq) + 2 Cr3+(aq) + 8 H2O(l) 2. What is the oxidation number of the UNDERLINED atom? (a) HCl (b) HOCl (c) S2O32– (d) As2O3 Na3AlCl6 (f) MnO42– (g) NH3 (h) Mg2P2O7 3. 4. (e) (a) -1 (b) +1 (c) +2 (d) +3 (e) +3 (f)+6 (g) -3 (h) +5 Write the formula for each of the following compounds: (a) potassium sulfide; K2S (b) barium chloride dihydrate; BaCl2.2H2O (c) antimony(III) sulfide; Sb2S3 (d) calcium hydride; CaH2 (e) potassium chlorate; KClO3 (f) sulfurous acid; H2SO3(aq) (g) calcium sulfite dihydrate; CaSO3.2H2O (h) chromium(III) oxide; Cr2O3 Which of the following reactions is an oxidation-reduction (REDOX) reaction. Write the NET-IONIC equation for each reaction. Indicate where applicable, (i) the species being oxidized and the species being reduced; and (ii) the oxidant (oxidizing agent) and the reductant (reducing agent). If REDOX reaction : Cu being reduced/oxidant S being oxidized/reductant (a) 2 Cu2O + Cu2S → 6 Cu + SO2 2 Cu2O + Cu2S → 6 Cu + SO2 (b) AgNO3(aq) + NaCl(aq) → AgCl(aq) + NaNO3(aq) Not redox. Double replacement reaction (c) 2 Al(s) + 3 F2(g) → 2 AlF3(s) 2 Al(s) + 3 F2(g) → 2 AlF3(s) (d) 2 K2CrO4(aq) + H2SO4(aq) → K2Cr2O7(aq) + K2SO4(aq) Not redox (e) Cl2(aq) + 2 KI(aq) → I2(aq) + 2 KCl(aq) Cl2(aq) + 2 KI(aq) → I2(aq) + 2 KCl(aq) Na2O(s) + H2O(l) → 2 NaOH(aq) Not redox (g) 2 CuS(s) + 3 O2(g) → 2 CuO(s) + 2 SO2(g) 2 CuS(s) + 3 O2(g) → 2 CuO(s) + 2 SO2(g) Calculate the oxidation number of the indicated element in each of the following species. (f) 5. (a) (b) (c) (d) sulfur in thiosulfate ion; +2 nitrogen in ammonium chloride; -3 zinc in Na2ZnO2 ; +2 uranium in UO2(NO3)2; +2 Chemistry 1010 Loader (Fall 2004) 6. (e) silicon in silicic acid (like carbonic acid); +4 (f) phosphorus in the fertilizer called triple superphosphate, Ca(H2PO4)2; +5 Write balanced half-reactions for each of the following partial half reactions. (a) ClO2–(aq) → Cl2(g) in ACID solution 6 e– + 8 H+(aq) + 2 ClO2–(aq) → Cl2(g) + 4 H2O(l) (b) S2O32– → S4O62– in ACID solution 2 S2O32– → S4O62– + 2 e– (c) Cr(OH)3 → CrO42– in BASIC solution 5 OH – + Cr(OH)3 → CrO42– + 3 e– + 4 H2O(l) 7. (d) N2H4 → N2 in BASIC solution 4 OH – + N2H4 → N2 + 4 e – + 4 H2O(l) Balance the following redox equations in acid solution. (a) MnO2(s) + PbO2(aq) → MnO4–(aq) + Pb2+(aq) reduction PbO2(aq) → Pb2+(aq) oxidation MnO2(s) → MnO4–(aq) – 2 e + PbO2(aq) → Pb2+(aq) change in oxidation number = -2 MnO2(s) → MnO4–(aq) + 3 e– change in oxidation number = +3 2 MnO2(s) + 3 Pb–O2(aq) → 2 MnO4–(aq) + 3 Pb2+(aq) :balance change in ON 4 H+(aq) + 2 MnO2(s) + 3 PbO2(aq) → 2 MnO4–(aq) + 3 Pb2+(aq) :balance charges with H+ 4 H+(aq) + 2 MnO2(s) + 3 PbO2(aq) → 2 MnO4–(aq) + 3 Pb2+(aq) + 2 H2O(l) :balance H with H2O BALANCED: 4 H+(aq) + 2 MnO2(s) + 3 PbO2(aq) → 2 MnO4–(aq) + 3 Pb2+(aq) + 2 H2O(l) (b) Br2(aq) + SO2(g) → H2SO4(aq) + HBr(aq) 2 H2O(l) + Br2(aq) + SO2(g) → H2SO4(aq) + 2 HBr(aq) 8. (d) Cr2O72–(aq) + I–(aq) → Cr3+(aq) + I2(s) (acid solution) 14 H+(aq) + Cr2O72–(aq) + 6 I–(aq) → 2 Cr3+(aq) + I2(s) + 7 H2O(l) Concentrated sulfuric acid has a density of 1.84 g.cm–3 (1.00 mL = 1.00 cm3) and contains 96% of H2SO4 by weight. (a) Calculate the concentration of the concentrated sulfuric acid in mol.L–1. Ans: 18.0 mol L–1 (b) What volume of the concentrated sulfuric acid would be required to make 2.50 litres of a 3.0 mol.L–1 solution of the bench acid? Ans: 0.417 L (c) What is the mass of the volume of the acid calculated in (b)? Ans: 0.767 kg 9. (a) Calculate the concentration in mol.L–1 of a sodium hydroxide solution 33.45 mL of which neutralizes 25.00 mL of 0.1500 mol.L–1 of nitric acid. NaOH(aq) + HNO3(aq) → NaNO3(aq) + H2O(l) mol HNO3 neutralized = 25.00 x 10–3 L x 0.1500 mol.L–1 = 3.750 x 10–3 mol. Since one mole of nitric acid reacts with one mole of sodium hydroxide.. mole NaOH = 3.750 x 10–3 mol concentration of NaOH = 0.1121 mol.L–1 (b) A 10.00 mL aliquot of dilute sulfuric acid required 16.12 mL of 0.1021 mol.L–1 sodium hydroxide solution for neutralization. What was the concentration of the sulfuric acid solution? 2 NaOH(aq) + H2SO4(aq) → Na2SO4 (aq) + 2 H2O(l) NaOH used = 16.12 x 10–3 L x 0.1021 mol.L–1 = 1.6459 x 10–3 mol Since one mole of sulfuric acid reacts with two moles of sodium hydroxide. mole NaOH = 1 2 x 1.6459 x 10–3 mol = 8.2293 x 10–4 mol Page 2 of 2 concentration of sulfuric acid = 0.08229 mol.L–1 (c) A 0.64753 g sample of the hydroxide of an unknown metal (M) required 34.30 mL of 0.4244 hydrochloric acid for complete reaction. The hydroxide is insoluble in water and may have the formula MOH, M(OH)2 or M(OH)3. Use the information from the titration to deduce the most likely formula and hence the molar mass of the metal. The possible reactions are MOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) M(OH)2(aq) + 2 HCl(aq) → 2 NaCl(aq) + 2 H2O(l) M(OH)3(aq) + 3 HCl(aq) → 3 NaCl(aq) + 3 H2O(l) mol HCl used in the reaction = 34.30 x 10–3 L x 0.4244 mol.L–1 = 0.014557 mol For the first equation one mole of acid reacts with one mole of base so if MOH is the formula then the molar mass would be 0.64753 g -1 0.014557 mol = 44.482 g mol giving about 44.48 - 17.00 = 37.48 g mol-1 for the molar mass of M for M(OH)2(aq) the molar mass would be 2 x 44.482 g mol-1 = 88.965 g mol-1 giving about 89.87 - 34.00 = 55.87 g mol-1 for the molar mass of M. for M(OH)3(aq) the molar mass would be 3 x 44.482 g mol-1 = 133.45 g mol-1 giving about 133.45 - 51.00 = 82.45 g mol-1 for the molar mass of M. The is no metal with a molar mass close to 37.48 or 82.45 g mol-1 with an insoluble hydroxide. However, for 55.87 g mol-1 this is almost exactly the molar mass of iron (55.85 g mol-1) so the metal hydroxide is probably Fe(OH)2. 10. A 50.00 mL sample of a solution containing iron(II) ions, Fe2+(aq), required 20.63 mL of a 0.0216 mol.L–1 solution of KMnO4 for complete reaction according to the equation: (incorrect wording in the original) 8 H+(aq) + MnO4–(aq) + 5 Fe2+(aq) → Mn2+(aq) + 5 Fe3+(aq) + 4 H2O(l) (a) Calculate the concentration of iron(II) ions in the solution. 0.04456 mol L-1 (b) What volume of a 0.0150 mol.L–1 solution of KMnO4 solution would be required to react completely with the iron(II) sulfate solution formed when 0.23245 g of iron wire reacted with sulfuric acid? Fe(s) + H2SO4(aq) → FeSO4(aq) + H2(g) mol iron = 0.23245 g = 4.1620 x 10-3 mol 55.85 g mol -1 1 mol KMnO4 required = 5 x 4.1620 x 10-3 mol 1 -3 5 x 4.1620 x10 mol so volume required = = 5.549 x 10-–3 L 0.150 mol L -1 Ans: 5.549 mL 11. Balance the following equations in acid or base as indicated. Identify the oxidant and the reductant. (a) ClO3– + MnO4– → MnO2 + ClO4– (acidic) 2 H+ + 3 ClO3– + 2 MnO4– → 2 MnO2 + 3 ClO4– + H2O (b) NO3– + S → NO + H2SO3 (acidic) 4 H+ + 4 NO3– + 3 S + H2O → 4 NO + 3 H2SO3 (c) Cr2O72– + I2 → IO3– + Cr3+ (acidic) 34 H+ + 5 Cr2O72– + 3 I2 → 6 IO3– + 10 Cr3+ + 17 H2O (d) Zn + NO3– → ZnO22– + NH3 (basic) 7 OH–+ 4 Zn + NO3– → 4 ZnO22– + NH3 + 2 H2O (e) S2– + ClO3– → Cl– + S (basic) 3 H2O + 3 S2– + ClO3– → Cl– + 3 S + 6 OH–