Firstly, I'd like to so thank you for all your help so far =D I have an chemistry assignment due tomorrow, and just wanted to double check some questions. Questions 1. (a) Write balanced molecular equations for the following potential precipitation reactions. Indicate the states of reactants and products [(aq) or (s)]. (b) In those cases where a precipitate forms, write the net ionic equation. If there is no reaction, state "No reaction." (i) NH4Cl (aq) + Ni(NO3)2 (aq) (ii) (NH4)2SO4 (aq) + Pb(NO3)2 (aq) (iii) KOH (aq) + Fe(CH3CO2)2 (aq) (iv) Na2CO3 (aq) + Fe(ClO4)2 (aq) 2. Each of the following salts can be prepared from an acid and a base. Write the balanced molecular equation and the net ionic equation for the preparation of each. Indicate states of the reactants and products [(aq), (s), (l)]. Review solubility rules if necessary to determine the solubility of the reactants. (a) KNO3 (aq) (b) NaHCO3 (aq) (c) MnCl2 (aq) 3. Assume that 21.9675 g of (NH4)2SO4 is dissolved in enough water to make 500.0 mL of solution. (a) What is the molarity of (NH4)2SO4? (b) What is the molarity of the ammonium cation? (c) What is the molarity of the sulfate anion? The molar mass of (NH4)2SO4 is 132.140 g/mol. 4. You have a solution of 0.1609 M zinc(II) chloride. (a) How many moles of solute are contained in 47.13 mL of solution? (b) How many grams of solute are contained in 47.13 mL of solution? (c) What volume (in millilitres) of solution is needed to obtain 0.0133 moles of solute? 5. Suppose you need 2.50 L of 1.50 M acetic acid, but only the concentrated acid (17.4 M) is available. What volume (in L) of the concentrated acid must you dilute to 2.50 L to obtain 1.50 M acetic acid? 6. Calcium carbonate reacts with hydrochloric acid to produce calcium chloride, carbon dioxide, and water. How many grams of calcium carbonate are required for complete reaction with 35.0 mL of 0.461 M hydrochloric acid? The molar mass of calcium carbonate is 100.087 g/mol. 7. A sample contains an unknown amount of the mineral calcite, CaCO3. If 0.6091 g of the sample requires 40.71 mL of 0.1000 M HCl to neutralize the CaCO3 completely, what is the percentage of CaCO3 in the sample? The molar mass of CaCO3 is 100.09 g/mol. The balanced equation for the reaction is: CaCO3 + 2 HCl -> H2O + CO2 + CaCl28. Sodium thiosulfate, Na2S2O3, is an important reagent for titrations. Its solutions can be standardized by titrating the iodine released when a weighed amount of potassium hydrogen iodate, KH(IO3)2 (389.912g/mol), is allowed to react with excess potassium iodide in acidic solutions. The net ionic equations are: production of iodine from KH(IO3)2: IO3 + 5I + 6H -> 3I2 + 3H2O titration of iodine: I2 + 2 S2O32 -> 2I + S4O62 What is the molarity of a sodium thiosulfate solution if 27.62 mL are required to titrate the iodine released from 0.1695 g of KH(IO3)2? 9. You have 0.6395 g of an unknown monoprotic acid, HA, which reacts with NaOH according to the balanced equation HA + NaOH NaA + H2O If 39.24 mL of 0.1041 M NaOH is required to titrate the acid to the equivalence point, what is the molar mass of the acid? Answers 1. a) 2NH4CL (Q) + Ni(NO3)2 -> Ni(NH4)2(aq) + 2CLNO3(AQ) NR b) (NH4)2SO4 (AQ) + Pb(NO3)2 -> 2NH4NO3 (AQ) + PbSO4 (AQ) NR c) 2KOH + FE(CH3CO2) -> K2(CHCCO2) + Fe(OH)2 (S) NIE: 2OH + FE -> FE(OH)2 d) Na2CO3 + Fe(ClO4)2 -> Na2(ClO4)2 + FeCo3 (S) 2. A) HNO3 + KOH -> KNO3 + H20 H + OH -> H20 B) H2CO3 + NaOH -> HCO3 + H20 H2CO3 + OH -> HCO3 + H20 C) 2HCl + Mn(OH)2 -> 2MnCl2 + 2H20 2H + Mn(OH)2 -> Mn + H20 3. A) 21.9675 / 132.140 = 0.166244 mol 0.166244 mol / 0.500 L = 0.332 B) NH4 = 14.0067 + (4 X 1.008) = 18.04 g/mol 2 MOL NH4 PER 1 MOL (NH4)2SO4 M = 0.332 / 0.500 = 0.665 4. A) 0.1609 = X / 0.04713 0.1609 x 0.04713 = X X = 0.007583 B) g = 0.007583 x 136.2954 = 1.034g C) 0.007583 / 0.0133 = 1.75 47.13 / 1.75 = 82.51 5. 17.4 M = 17.5 mol / 1L 1.5 M = 1.5 mol / 1L 17.4 / 1.5 = 11.6L 17.4 mol / 11.6L = 1.5 M 11.6L are needed 6. CaCO3 + 2HCl -> CaCl2 + CO2 + H2O 0.461 M = X / 0.0350 0.461 X 0.0350 = X X = 0.016 mol HCL 2 mol H2O per mole of CaCO3, so mol CaCO3 = 0.016 / 2 = 0.0080 mol 0.0080 X 100.087g per mol = 0.81g CaCO3 7. 0.1000 M = 1 mol / 10L 2 moles HCl per 1 mole CaCO3 0.1000 x 0.4071 = 0.004071 mol HCl 0.004071 / 2 x 100.078g = 0.2037g 0.2037 / 0.6091 x 100 = 33.44% of the mineral is calcite 8. 6 mol I- per 1 mol KH(IO3) 0.1695g/mol / 389.912g = 0.0004347mol Na2S2O3 0.004347 mol / 0.02762 = 0.01574 M = 0.01574 9. 0.1041 x 0.03924 = 0.004085 mol NaOH thus there is 0.004085 mol HA 0.6395g / 0.004085 = 156.6 g/mol Sorry, I know it's long. I'm just looking for big glaring errors M.