CHAPTER TYPES OF CHEMICA REACTIONS AND SOUTION STOICHIOMETRY Questions 17. a. Polarity is a term applied to covalent compounds. Polar covalent compounds have an unequal sharing of electrons in bonds that results in unequal charge distribution in the overall molecule. Polar molecules have a partial negative end and a partial positive end. These are not full charges as in ionic compounds but are charges much smaller in magnitude. Water is a polar molecule and dissolves other polar solutes readily. The oxygen end of water (the partial negative end of the polar water molecule) aligns with the partial positive end of the polar solute, whereas the hydrogens of water (the partial positive end of the polar water molecule) align with the partial negative end of the solute. These opposite charge attractions stabilize polar solutes in water. This process is called hydration. Nonpolar solutes do not have permanent partial negative and partial positive ends; nonpolar solutes are not stabilized in water and do not dissolve. b. KF is a soluble ionic compound, so it is a strong electrolyte. KF(aq) actually exists as separate hydrated K ions and hydrated F ions in solution: C 6H 1O 6 is a polar covalent molecule that is a nonelectrolyte. C 6H 1O 6 is hydrated as described in part a. c. RbCl is a soluble ionic compound, so it exists as separate hydrated Rb ions and hydrated Cl ions in solution. AgCl is an insoluble ionic compound, so the ions stay together in solution and fall to the bottom of the container as a precipitate. d. HNO is a strong acid and exists as separate hydrated H ions and hydrated NO ions in solution. CO is a polar covalent molecule and is hydrated as explained in part a. 18..0.0 mol/ 6.0 mol HCl; the.0 of solution contains 6.0 mol of the solute. HCl is a strong acid; it exists in aqueous solution as separate hydrated H ions and hydrated Cl ions. So the solution will contain 6.0 mol of H (aq) and 6.0 mol of Cl (aq). For the acetic acid solution, HC H O is a weak acid instead of a strong acid. Only some of the 6.0 moles of HC H O molecules will dissociate into H (aq) and C H O (aq). The.0 of.0 M HC H O solution will contain mostly hydrated HC H O molecules but will also contain some hydrated H ions and hydrated C H O ions. 19. Only statement b is true. A concentrated solution can also contain a nonelectrolyte dissolved in water, e.g., concentrated sugar water. Acids are either strong or weak electrolytes. Some ionic compounds are not soluble in water, so they are not labeled as a specific type of electrolyte. 100
CHAPTER SOUTION STOICHIOMETRY 101 0. One mole of NaOH dissolved in 1.00 of solution will produce 1.00 M NaOH. First, weigh out 0.00 g of NaOH (1.000 mol). Next, add some water to a 1- volumetric flask (an instrument that is precise to 1.000 ). Dissolve the NaOH in the flask, add some more water, mix, add more water, mix, etc. until water has been added to 1.000- mark of the volumetric flask. The result is 1.000 of a 1.000 M NaOH solution. Because we know the volume to four significant figures as well as the mass, the molarity will be known to four significant figures. This is good practice, if you need a three-significant-figure molarity, your measurements should be taken to four significant figures. When you need to dilute a more concentrated solution with water to prepare a solution, again make all measurements to four significant figures to ensure three significant figures in the molarity. Here, we need to cut the molarity in half from.00 M to 1.00 M. We would start with 1 mole of NaOH from the concentrated solution. This would be 500.0 m of.00 M NaOH. Add this to a 1- volumetric flask with addition of more water and mixing until the 1.000- mark is reached. The resulting solution would be 1.00 M. 1. Use the solubility rules in Table.1. Some soluble bromides by Rule would be NaBr, KBr, and NH Br (there are others). The insoluble bromides by Rule would be AgBr, PbBr, and Hg Br. Similar reasoning is used for the other parts to this problem. Sulfates: Na SO, K SO, and (NH ) SO (and others) would be soluble, and BaSO, CaSO, and PbSO (or Hg SO ) would be insoluble. Hydroxides: NaOH, KOH, Ca(OH) (and others) would be soluble, and Al(OH), Fe(OH), and Cu(OH) (and others) would be insoluble. Phosphates: Na PO, K PO, (NH ) PO (and others) would be soluble, and Ag PO, Ca (PO ), and FePO (and others) would be insoluble. ead: PbCl, PbBr, PbI, Pb(OH), PbSO, and PbS (and others) would be insoluble. Pb(NO ) would be a soluble Pb salt.. Pb(NO ) (aq) KI(aq) PbI (s) KNO (aq) (formula equation) Pb (aq) NO (aq) K (aq) I (aq) PbI (s) K (aq) NO (aq) (complete ionic equation) The 1.0 mol of Pb ions would react with the.0 mol of I ions to form 1.0 mol of the PbI precipitate. Even though the Pb and I ions are removed, the spectator ions K and NO are still present. The solution above the precipitate will conduct electricity because there are plenty of charge carriers present in solution.. The Brønsted-owry definitions are best for our purposes. An acid is a proton donor, and a base is a proton acceptor. A proton is an H ion. Neutral hydrogen has 1 electron and 1 proton, so an H ion is just a proton. An acid-base reaction is the transfer of an H ion (a proton) from an acid to a base.. The acid is a diprotic acid (H A), meaning that it has two H ions in the formula to donate to a base. The reaction is H A(aq) NaOH(aq) H O(l) Na A(aq), where A is what is left over from the acid formula when the two protons (H ions) are reacted.
10 CHAPTER SOUTION STOICHIOMETRY For the HCl reaction, the base has the ability to accept two protons. The most common examples are Ca(OH), Sr(OH), and Ba(OH). A possible reaction would be HCl(aq) Ca(OH) (aq) H O(l) CaCl (aq). 5. a. The species reduced is the element that gains electrons. The reducing agent causes reduction to occur by itself being oxidized. The reducing agent generally refers to the entire formula of the compound/ion that contains the element oxidized. b. The species oxidized is the element that loses electrons. The oxidizing agent causes oxidation to occur by itself being reduced. The oxidizing agent generally refers to the entire formula of the compound/ion that contains the element reduced. c. For simple binary ionic compounds, the actual charges on the ions are the same as the oxidation states. For covalent compounds, nonzero oxidation states are imaginary charges the elements would have if they were held together by ionic bonds (assuming the bond is between two different nonmetals). Nonzero oxidation states for elements in covalent compounds are not actual charges. Oxidation states for covalent compounds are a bookkeeping method to keep track of electrons in a reaction. 6. Mass balance indicates that we have the same number and type of atoms on both sides of the equation (so that mass is conserved). Similarly, net charge must also be conserved. We cannot have a buildup of charge on one side of the reaction or the other. In redox equations, electrons are used to balance the net charge between reactants and products. Exercises Aqueous Solutions: Strong and Weak Electrolytes 7. a. NaBr(s) Na (aq) Br - (aq) b. MgCl (s) Mg (aq) Cl (aq) Na Br - Br - Na Na Br - Your drawing should show equal number of Na and Br - ions. Cl - Mg Mg Cl - Cl - Cl - Cl - Cl - Mg Your drawing should show twice the number of Cl ions as Mg ions. c. Al(NO ) (s) Al (aq) NO (aq) d. (NH ) SO (s) NH (aq) SO (aq) Al - NO - NO - NO - NO - NO - NO Al - NO Al - NO - NO - SO NH - SO NH NH NH SO - NH NH
CHAPTER SOUTION STOICHIOMETRY 10 For e-i, your drawings should show equal numbers of the cations and anions present because each salt is a 1 : 1 salt. The ions present are listed in the following dissolution reactions. e. NaOH(s) Na (aq) OH (aq) f. FeSO (s) Fe (aq) SO (aq) g. KMnO (s) K (aq) MnO (aq) h. HClO (aq) H (aq) ClO (aq) i. NH C H O (s) NH (aq) C H O (aq) 8. a. Ba(NO ) (aq) Ba (aq) NO (aq); picture iv represents the Ba and NO ions present in Ba(NO ) (aq). b. NaCl(aq) Na (aq) Cl (aq); picture ii represents NaCl(aq). c. K CO (aq) K (aq) CO (aq); picture iii represents K CO (aq). d. MgSO (aq) Mg (aq) SO (aq); picture i represents MgSO (aq). HNO (aq) H (aq) NO (aq). Picture ii best represents the strong acid HNO. Strong acids are strong electrolytes. HC H O only partially dissociates in water; acetic acid is a weak electrolyte. None of the pictures represent weak electrolyte solutions; they all are representations of strong electrolytes. 9. CaCl (s) Ca (aq) Cl (aq) 0. MgSO (s) Mg (aq) SO (aq); NH NO (s) NH (aq) NO (aq) Solution Concentration: Molarity 1. a. 5.6 g NaHCO 1 mol NaHCO 8.01 g NaHCO 6.69 10 mol NaHCO M 6.69 10 mol 50.0 m 1000 m 0.677 M NaHCO b. 0.186 g K Cr O 7 1 mol K CrO 7 9.0 g K Cr O 7 6.75 10 mol K Cr O 7 M 6.75 10 500.0 10 mol 1.55 10 M K Cr O 7 c. 0.105 g Cu 1 mol Cu 6.55 g Cu 1.61 10 mol Cu 1.61 10 mol Cu M 1.61 10 mol Cu 00.0 m 1000 m 8.065 10 M Cu
10 CHAPTER SOUTION STOICHIOMETRY. 75.0 m 0.79 g m 1 mol 1. mol C H 5OH; molarity 6.07 g 1. mol 0.50 5. M C H 5OH. a. M Ca(NO ) 0.100 mol Ca(NO ) 1.00 M 0.100 Ca(NO ) (s) Ca (aq) NO (aq); M 1.00 M; M (1.00).00 M Ca NO b. M Na SO.5 mol Na SO 1.5.0 M Na SO (s) Na (aq) SO (aq); M (.0).0 M ; M Na SO.0 M c. 5.00 g NH Cl 1 mol NH Cl 0.095 mol NH Cl 5.9 g NH Cl M NH Cl 0.095 mol NH Cl 0.187 M 0.5000 NH Cl(s) NH (aq) Cl (aq); M M NH Cl 0.187 M d. 1.00 g K PO mol K PO.71 10 mol K PO 1.7 g 1 M K PO.71 10 mol 0.500 0.0188 M K PO (s) K (aq) PO (aq); M (0.0188) 0.056 M; M K PO 0.0188 M. a. M Na PO 0.000 mol 0.0100.00 M Na PO (s) Na (aq) PO (aq) M (.00) 6.00 M; M.00 M ; Na PO b. M Ba(NO ) 0.00 mol 0.6000 0.500 M Ba(NO ) (s) Ba (aq) NO (aq); M 0.500 M; M (0.500) 1.00 M Ba NO c. M KCl 1 mol KCl 1.00 g KCl 7.55 g KCl 0.5000 0.068 M KCl(s) K (aq) Cl (aq); M M K Cl 0.068 M
CHAPTER SOUTION STOICHIOMETRY 105 d. M (NH ) SO 1 mol (NH) SO 1 g (NH) SO 1.15 g 1.50 0.666 M (NH ) SO (s) NH (aq) SO (aq) M (0.666) 1. M; M 0.666 M NH SO 5. Mol solute volume () molarity mol ; AlCl (s) Al (aq) Cl (aq) Mol Cl 0.1000 0.0 mol AlCl mol Cl 9.0 10 mol Cl mol AlCl MgCl (s) Mg (aq) Cl (aq) Mol Cl 0.60 mol MgCl 0.0500 mol Cl 6.0 10 mol Cl mol MgCl NaCl(s) Na (aq) Cl (aq) Mol Cl 0.0 mol NaCl 0.000 1 mol Cl 8.0 10 mol Cl mol NaCl 100.0 m of 0.0 M AlCl contains the most moles of Cl ions. 6. NaOH(s) Na (aq) OH (aq), total mol of ions (1 mol Na and 1 mol Cl ) per mol NaOH. 0.1000 0.100 mol NaOH mol ions.0 10 mol ions mol NaOH BaCl (s) Ba (aq) Cl (aq), total mol of ions per mol BaCl. 0.0500 0.00 mol mol ions.0 10 mol ions mol BaCl Na PO (s) Na (aq) PO (aq), total mol of ions per mol Na PO. 0.150 mol Na PO 0.0750 mol ions.50 10 mol ions mol Na PO 75.0 m of 0.150 M Na PO contains the largest number of ions. 7. Molar mass of NaOH.99 16.00 1.008 0.00 g/mol 0.00 mol NaOH 0.00 g NaOH Mass NaOH 0.500.00 g NaOH mol NaOH
106 CHAPTER SOUTION STOICHIOMETRY 8. 10. g AgNO 1 mol AgNO 169.9 g 1 0. 0 m 0.5 mol AgNO 9. 0.0150 17 mmol Na 1mol.99 g Na 0.07 g Na 1000 mmol mol Na 0. 0.00 g C 7H 6O 0.059 g C 7H 6O 1 mol C 7 H 6O 1.0 10 mol C 7H 6O 86.6 g C H O 7 6 1 mol C 7 H 6O 1.5 10 mol C 7H 6O 86.6 g C H O 7 6 1.0 10 mol 1.0 10 1 M; 1d 10 d 1.5 10 mol 1.5 10 1 M 1d 10 d HD levels between 0.59 mg/d correspond to a molarity of cholesterol between 1.0 10 and 1.5 10 M. 1. a..00 0.50 mol NaOH 0.00 g NaOH 0.0 g NaOH mol NaOH Place 0.0 g NaOH in a - volumetric flask; add water to dissolve the NaOH, and fill to the mark with water, mixing several times along the way. b..00 0.50 mol NaOH 1 stock 0.500 1.00 mol NaOH Add 500. m of 1.00 M NaOH stock solution to a - volumetric flask; fill to the mark with water, mixing several times along the way. 0.100 mol K CrO 19.0 g K CrO c..00 8.8 g K CrO mol K CrO Similar to the solution made in part a, instead using 8.8 g K CrO. 0.100 mol K CrO d..00 1 stock 0.11 1.75 mol K CrO Similar to the solution made in part b, instead using 11 m of the 1.75 M K CrO stock solution.. a. 1.00 solution.50 mol H SO 0.50 mol H SO 0 0.50 mol H SO 1 18 mol H SO.8 10 conc. H SO or 8 m
CHAPTER SOUTION STOICHIOMETRY 107 Dilute 8 m of concentrated H SO to a total volume of 1.00 with water. The resulting 1.00 of solution will be a 0.50 M H SO solution. b. We will need 0.50 mol HCl. 0.50 mol HCl 1 1 mol HCl. 10 m Dilute m of concentrated HCl to a final volume of 1.00. c. We need 0.50 mol NiCl. 1 mol NiCl 6H O 0.50 mol NiCl mol NiCl 7.69 g NiCl 6H O mol NiCl 6H O 118.8 g NiCl 6H O 10 g Dissolve 10 g NiCl 6H O in water, and add water until the total volume of the solution is 1.00. 0.50 mol HNO d. 1.00 0.50 mol HNO 0.50 mol HNO 1 16 mol HNO 0.01 1 m Dissolve 1 m of concentrated reagent in water. Dilute to a total volume of 1.00. e. We need 0.50 mol Na CO. 105.99 g Na CO 0.50 mol Na CO mol 5 g Na CO Dissolve 5 g Na CO in water, dilute to 1.00.. 10.8 g (NH ) SO 1 mol 1.15 g 8.17 10 mol (NH ) SO Molarity 8.17 10 mol 100.0 m 1000 m 0.817 M (NH ) SO Moles of (NH ) SO in final solution: 10.00 10 0.817 mol Molarity of final solution 8.17 10 mol 8.17 10 mol (10.00 50.00) m 1000 m 0.16 M (NH ) SO (NH ) SO (s) NH (aq) SO (aq); M (0.16) 0.7 M; M 0.16 M NH SO
108 CHAPTER SOUTION STOICHIOMETRY. Molarity total mol HNO total volume ; total volume 0.05000 0.10000 0.15000 0.100 mol HNO Total mol HNO 0.05000 0.10000 0.00 mol HNO Total mol HNO 5.00 10 mol.00 10 mol.50 10 mol HNO Molarity.50 10 mol HNO 0.15000 0.167 M HNO As expected, the molarity of HNO is between 0.100 M and 0.00 M. 5. Mol Na CO 0.0700.0 mol Na CO 0.1 mol Na CO Na CO (s) Na (aq) CO (aq); mol Na (0.1 mol) 0. mol 1.0 mol NaHCO Mol NaHCO 0.000 0.00 mol NaHCO NaHCO (s) Na (aq) HCO (aq); mol Na 0.00 mol M Na total mol Na total volume 0. mol 0.00 mol 0.0700 0.000 0.5 mol 0.1000.5 M Na 0.50 mol CoCl 6. Mol CoCl 0.0500 0.015 mol Mol NiCl 0.050 0.50 mol NiCl 0.00875 mol Both CoCl and NiCl are soluble chloride salts by the solubility rules. A 0.015-mol aqueous sample of CoCl is actually 0.015 mol Co and (0.015 mol) 0.050 mol Cl. A 0.00875- mol aqueous sample of NiCl is actually 0.00875 mol Ni and (0.00875) 0.0175 mol Cl. The total volume of solution that these ions are in is 0.0500 0.050 0.0750. 0.015 mol Co 0.00875 mol Ni M 0.167 M ; M 0.117 M Co Ni 0.0750 0.0750 0.050 mol Cl 0.0175 mol Cl M 0. 567 Cl 0.0750 M 7. Stock solution 10.0 mg 500.0 m 10.0 10 g 500.0 m 5.00 10 g steroid m 100.0 10 6 stock 5 1000 m.00 10 g steroid.00 10 6 g steroid m
CHAPTER SOUTION STOICHIOMETRY 109 This is diluted to a final volume of 100.0 m..00 6 10 g steroid 100.0 m 1000 m 1 mol steroid 6. g steroid 5.9 10 8 M steroid 8. Stock solution: 1.58 g Mn Molarity 50.00 m Molarity Solution B: 10.0 m Molarity Solution C: 1 mol Mn 5.9 g Mn.8 10 mol Mn 1.000 1 1000 m 1. 10 mol 1000.0 m 1 1000 m 10.00 10 Molarity.88 10 mol Mn.88 10 M Solution A:.8 10 mol 1. 10 mol Mn 1000 m 1 1. 10 M 1. 10 mol 1. 10 5 mol Mn 5 1. 10 mol 0.500 5.768 10 7 5 5.768 10 mol 0.5000 5.768 10 5 M mol 5.768 10 7 mol Mn 1.15 10 6 M Precipitation Reactions 9. The solubility rules referenced in the following answers are outlined in Table.1 of the text. a. Soluble: Most nitrate salts are soluble (Rule 1). b. Soluble: Most chloride salts are soluble except for Ag, Pb, and Hg (Rule ). c. Soluble: Most sulfate salts are soluble except for BaSO, PbSO, Hg SO, and CaSO (Rule.) d. Insoluble: Most hydroxide salts are only slightly soluble (Rule 5). Note: We will interpret the phrase slightly soluble as meaning insoluble and the phrase marginally soluble as meaning soluble. So the marginally soluble hydroxides Ba(OH), Sr(OH), and Ca(OH) will be assumed soluble unless noted otherwise.
110 CHAPTER SOUTION STOICHIOMETRY e. Insoluble: Most sulfide salts are only slightly soluble (Rule 6). Again, slightly soluble is interpreted as insoluble in problems like these. f. Insoluble: Rule 5 (see answer d). g. Insoluble: Most phosphate salts are only slightly soluble (Rule 6). 50. The solubility rules referenced in the following answers are from Table.1 of the text. The phrase slightly soluble is interpreted to mean insoluble, and the phrase marginally soluble is interpreted to mean soluble. a. Soluble (Rule ) b. Soluble (Rule 1) c. Insoluble (Rule ) d. Soluble (Rules and ) e. Insoluble (Rule 6) f. Insoluble (Rule 5) g. Insoluble (Rule 6) h. Soluble (Rule ) 51. In these reactions, soluble ionic compounds are mixed together. To predict the precipitate, switch the anions and cations in the two reactant compounds to predict possible products; then use the solubility rules in Table.1 to predict if any of these possible products are insoluble (are the precipitate). Note that the phrase slightly soluble in Table.1 is interpreted to mean insoluble, and the phrase marginally soluble is interpreted to mean soluble. a. Possible products FeCl and K SO ; both salts are soluble, so no precipitate forms. b. Possible products Al(OH) and Ba(NO ) ; precipitate Al(OH) (s) c. Possible products CaSO and NaCl; precipitate CaSO (s) d. Possible products KNO and NiS; precipitate NiS(s) 5. Use Table.1 to predict the solubility of the possible products. a. Possible products Hg SO and Cu(NO ) ; precipitate Hg SO b. Possible products NiCl and Ca(NO ) ; both salts are soluble so no precipitate forms. c. Possible products KI and MgCO ; precipitate MgCO d. Possible products NaBr and Al (CrO ) ; precipitate Al (CrO ) 5. For the following answers, the balanced formula equation is first, followed by the complete ionic equation, then the net ionic equation. a. No reaction occurs since all possible products are soluble salts. b. Al(NO ) (aq) Ba(OH) (aq) Al(OH) (s) Ba(NO ) (aq) Al (aq) 6 NO (aq) Ba (aq) 6 OH (aq) Al(OH) (s) Ba (aq) 6 NO (aq) Al (aq) OH (aq) Al(OH) (s)
CHAPTER SOUTION STOICHIOMETRY 111 c. CaCl (aq) Na SO (aq) CaSO (s) NaCl(aq) Ca (aq) Cl (aq) Na (aq) SO (aq) CaSO (s) Na (aq) Cl (aq) Ca (aq) SO (aq) CaSO (s) d. K S(aq) Ni(NO ) (aq) KNO (aq) NiS(s) K (aq) S (aq) Ni (aq) NO (aq) K (aq) NO (aq) NiS(s) Ni (aq) S (aq) NiS(s) 5. a. Hg (NO ) (aq) CuSO (aq) Hg SO (s) Cu(NO ) (aq) Hg (aq) NO (aq) Cu (aq) SO (aq) Hg SO (s) Cu (aq) NO (aq) Hg (aq) SO (aq) Hg SO (s) b. No reaction occurs since both possible products are soluble. c. K CO (aq) MgI (aq) KI(aq) MgCO (s) K (aq) CO (aq) Mg (aq) I (aq) K (aq) I (aq) MgCO (s) Mg (aq) CO (aq) MgCO (s) d. Na CrO (aq) Al(Br) (aq) 6 NaBr(aq) Al (CrO ) (s) 6 Na (aq) CrO (aq) Al (aq) 6 Br (aq) 6 Na (aq) 6 Br (aq) Al (aq) CrO (aq) Al (CrO ) (s) Al (CrO ) (s) 55. a. When CuSO (aq) is added to Na S(aq), the precipitate that forms is CuS(s). Therefore, Na (the gray spheres) and SO (the bluish green spheres) are the spectator ions. CuSO (aq) Na S(aq) CuS(s) Na SO (aq); Cu (aq) S (aq) CuS(s) b. When CoCl (aq) is added to NaOH(aq), the precipitate that forms is Co(OH) (s). Therefore, Na (the gray spheres) and Cl - (the green spheres) are the spectator ions. CoCl (aq) NaOH(aq) Co(OH) (s) NaCl(aq) Co (aq) OH (aq) Co(OH) (s) c. When AgNO (aq) is added to KI(aq), the precipitate that forms is AgI(s). Therefore, K (the red spheres) and NO (the blue spheres) are the spectator ions. AgNO (aq) KI(aq) AgI(s) KNO (aq); Ag (aq) I (aq) AgI(s)
11 CHAPTER SOUTION STOICHIOMETRY 56. There are many acceptable choices for spectator ions. We will generally choose Na and NO as the spectator ions because sodium salts and nitrate salts are usually soluble in water. a. Fe(NO ) (aq) NaOH(aq) Fe(OH) (s) NaNO (aq) b. Hg (NO ) (aq) NaCl(aq) Hg Cl (s) NaNO (aq) c. Pb(NO ) (aq) Na SO (aq) PbSO (s) NaNO (aq) d. BaCl (aq) Na CrO (aq) BaCrO (s) NaCl(aq) 57. a. (NH ) SO (aq) Ba(NO ) (aq) NH NO (aq) BaSO (s) Ba (aq) SO (aq) BaSO (s) b. Pb(NO ) (aq) NaCl(aq) PbCl (s) NaNO (aq) Pb (aq) Cl (aq) PbCl (s) c. Potassium phosphate and sodium nitrate are both soluble in water. No reaction occurs. d. No reaction occurs because all possible products are soluble. e. CuCl (aq) NaOH(aq) Cu(OH) (s) NaCl(aq) Cu (aq) OH (aq) Cu(OH) (s) 58. a. CrCl (aq) NaOH(aq) Cr(OH) (s) NaCl(aq) Cr (aq) OH (aq) Cr(OH) (s) b. AgNO (aq) (NH ) CO (aq) Ag CO (s) NH NO (aq) Ag (aq) CO (aq) Ag CO (s) c. CuSO (aq) Hg (NO ) (aq) Cu(NO ) (aq) Hg SO (s) Hg (aq) SO (aq) Hg SO (s) d. No reaction occurs because all possible products (SrI and KNO ) are soluble. 59. Because a precipitate formed with Na SO, the possible cations are Ba, Pb, Hg, and Ca (from the solubility rules). Because no precipitate formed with KCl, Pb and Hg cannot be present. Because both Ba and Ca form soluble chlorides and soluble hydroxides, both these cations could be present. Therefore, the cations could be Ba and Ca (by the solubility rules in Table.1). For students who do a more rigorous study of solubility, Sr could also be a possible cation (it forms an insoluble sulfate salt, whereas the chloride and hydroxide salts of strontium are soluble).
CHAPTER SOUTION STOICHIOMETRY 11 60. Because no precipitates formed upon addition of NaCl or Na SO, we can conclude that Hg and Ba are not present in the sample because Hg Cl and BaSO are insoluble salts. However, Mn may be present since Mn does not form a precipitate with either NaCl or Na SO. A precipitate formed with NaOH; the solution must contain Mn because it forms a precipitate with OH [Mn(OH) (s)]. 61. AgNO (aq) Na CrO (aq) Ag CrO (s) NaNO (aq) 0.100 mol AgNO 0.0750 1 mol Na CrO 0.607 g Na CrO mol AgNO 161.98 g Na mol Na CrO CrO 6. Na PO (aq) Pb(NO ) (aq) Pb (PO ) (s) 6 NaNO (aq) 0.50 mol Pb(NO ) 0.1500 mol Na PO 0.50 mol Pb(NO ) 1 Na PO 0.100 mol Na PO 50. m Na PO 6. Fe(NO ) (aq) NaOH(aq) Fe(OH) (s) NaNO (aq) Assuming Fe(NO ) is limiting: 0.0750 0.105 mol Fe(NO ) Assuming NaOH is limiting: 0.150 mol NaOH 0.15 1 mol Fe(OH) mol Fe(NO ) 1 mol Fe(OH) mol NaOH 106.87 g Fe(OH) mol Fe(OH) 0.8 g Fe(OH) 106.87 g Fe(OH) 0.668 g Fe(OH) mol Fe(OH) Because NaOH produces the smaller mass of the Fe(OH) precipitate, NaOH is the limiting reagent and 0.668 g Fe(OH) can form. 6. The balanced equation is BaCl (aq) Fe (SO ) (aq) BaSO (s) FeCl (aq). 100.0 m BaCl 100.0 m Fe (SO ) 1 1000 m 1000 m 0.100 mol BaCl 0.100 mol Fe (SO 1 mol BaSO mol BaCl ). g BaSO mol BaSO. g BaSO mol BaSO mol Fe (SO ). g BaSO mol BaSO 7.00 g BaSO The BaCl reagent produces the smaller quantity of the BaSO precipitate, so BaCl is limiting and. g BaSO can form. 65. The reaction is AgNO (aq) NaBr(aq) AgBr(s) NaNO (aq). Assuming AgNO is limiting:
11 CHAPTER SOUTION STOICHIOMETRY 100.0 m AgNO 1 1000 m 0.150 mol AgNO AgNO 1 mol AgBr mol AgNO 187.8 g AgBr mol AgBr.8 g AgBr Assuming NaBr is limiting: 0.0 m NaBr 1 1000 m 1.00 mol NaBr NaBr 1 mol AgBr 187.8 g AgBr.76 g AgBr mol NaBr mol AgBr The AgNO reagent produces the smaller quantity of AgBr, so AgNO is limiting and.8 g AgBr can form. 66. AgNO (aq) CaCl (aq) AgCl(s) Ca(NO ) (aq) 0.0 mol AgNO 0.1000 mol AgCl 1. g AgCl.9 g AgCl mol AgNO mol AgCl 0.15 mol CaCl mol AgCl 1. g AgCl 0.1000. g AgCl mol CaCl mol AgCl AgNO is limiting (it produces the smaller mass of AgCl) and.9 g AgCl can form. The net ionic equation is Ag (aq) Cl (aq) AgCl(s). The ions remaining in solution are the unreacted Cl ions and the spectator ions NO and Ca (all Ag is used up in forming AgCl). The moles of each ion present initially (before reaction) can be easily determined from the moles of each reactant. We have 0.1000 (0.0 mol AgNO /) 0.00 mol AgNO, which dissolves to form 0.00 mol Ag and 0.00 mol NO. We also have 0.1000 (0.15 mol CaCl /) 0.015 mol CaCl, which dissolves to form 0.015 mol Ca and (0.015) 0.00 mol Cl. To form the.9 g of AgCl precipitate, 0.00 mol Ag will react with 0.00 mol of Cl to form 0.00 mol AgCl (which has a mass of.9 g). Mol unreacted Cl 0.00 mol Cl initially 0.00 mol Cl reacted Mol unreacted Cl 0.010 mol Cl M Cl 0.010 mol Cl total volume The molarities of the spectator ions are: 0.010 mol Cl 0.050 M Cl 0.1000 0.1000 0.00 mol NO 0.000 0.10 M NO ; 0.015 mol Ca 0.000 0.075 M Ca 67. a. The balanced formula equation is: KOH(aq) Mg(NO ) (aq) Mg(OH) (s) KNO (aq). b. The precipitate is magnesium hydroxide.
CHAPTER SOUTION STOICHIOMETRY 115 c. Assuming KOH is limiting: 0.1000 KOH 0.00 mol KOH KOH Assuming Mg(NO ) is limiting: 0.1000 Mg(NO ) 0.00 mol Mg(NO Mg(NO 1 mol Mg(OH) mol KOH ) ) 1 mol Mg(OH) mol Mg(NO 58. g Mg(OH) mol Mg(OH) 0.58 g Mg(OH) ) 58. g Mg(OH) 1.17 g Mg(OH) mol Mg(OH) The KOH reagent is limiting because it produces the smaller quantity of the Mg(OH) precipitate. So 0.58 g Mg(OH) can form. d. The net ionic equation for this reaction is Mg (aq) OH (aq) Mg(OH) (s). Because KOH is the limiting reagent, all of the OH is used up in the reaction. So M OH 0 M. Note that K is a spectator ion, so it is still present in solution after precipitation was complete. Also present will be the excess Mg and NO (the other spectator ion). Total Mg 0.1000 Mg(NO ) 0.00 mol Mg(NO Mg(NO ) ) 1 mol Mg mol Mg(NO ) 0.000 mol Mg Mol Mg reacted 0.1000 KOH 0.00 mol KOH 1 mol Mg(NO) KOH mol KOH 1 mol Mg 0.0100 mol Mg mol Mg(NO ) M Mg mol excess Mg total volume (0.000 0.0100) mol Mg 0.1000 0.1000 5.00 10 M Mg The spectator ions are K and NO. The moles of each are: mol K 0.1000 KOH 0.00 mol KOH 1 mol K 0.000 mol K KOH mol KOH mol NO 0.1000 Mg(NO ) The concentrations are: 0.00 mol Mg(NO Mg(NO ) ) mol NO mol Mg(NO ) 0.000 mol NO 0.000 mol K 0.100 M K ; 0.000 0.000 mol NO 0.000 0.00 M NO
116 CHAPTER SOUTION STOICHIOMETRY 68. a. Cu(NO ) (aq) KOH(aq) Cu(OH) (s) KNO (aq) Solution A contains.00.00 mol/.00 mol Cu(NO ), and solution B contains.00.00 mol/ 6.00 mol KOH. In the picture in the problem, we have formula units of Cu(NO ) ( Cu ions and 8 NO ions) and 6 formula units of KOH (6 K ions and 6 OH ions). With Cu ions and 6 OH ions present, OH is limiting (when all 6 molecules of OH react, we only need of the Cu ions to react with all of the OH present). After reaction, one Cu ion remains as Cu(OH) (s) formula units form as precipitate. The following drawing summarizes the ions that remain in solution and the relative amount of precipitate that forms. Note that K and NO ions are spectator ions. In the drawing, V 1 is the volume of solution A or B, and V is the volume of the combined solutions, with V V 1. The drawing exaggerates the amount of precipitate that would actually form. V NO - K NO - K NO - V 1 Cu K - - NO NO K NO - K NO - K NO - Cu(OH) Cu(OH) Cu(OH) b. The spectator ion concentrations will be one-half the original spectator ion concentrations in the individual beakers because the volume was doubled. Or using moles: M K 6.00 mol K 1.50 M and M NO.00 8.00 mol NO.00.00 M The concentration of OH ions will be zero because OH is the limiting reagent. From the drawing, the number of Cu ions will decrease by a factor of four as the precipitate forms. Because the volume of solution doubled, the concentration of Cu ions will decrease by a factor of eight after the two beakers are mixed: M.00 M Cu 1 0.50 M 8 Alternately, one could certainly use moles to solve for M : Cu Mol Cu.00 mol OH reacted.00 1 mol Cu mol OH.00 mol Cu reacted Mol Cu.00 mol Cu present initially.00.00 mol Cu present initially Excess Cu present after reaction.00 mol.00 mol 1.00 mol Cu excess
CHAPTER SOUTION STOICHIOMETRY 117 M Cu 1.00 mol Cu.00.00 0.50 M Mass of precipitate 6.00 mol KOH 69. M SO (aq) CaCl (aq) CaSO (s) MCl(aq) 1 mol CaSO 1.6 g CaSO 16.15 g CaSO 1 mol M 1 mol Cu(OH) mol KOH SO 9.99 10 mol M SO mol CaSO From the problem, 1. g M SO was reacted, so: molar mass 1. g M 9.99 10 SO mol M SO 1 g/mol 1 u (atomic mass M).07 (16.00), atomic mass M u From periodic table, M Na (sodium). 97.57 g Cu(OH) mol Cu(OH) 9 g Cu(OH) 70. a. Na, NO, Cl, and Ag ions are present before any reaction occurs. The excess Ag added will remove all of the Cl ions present. Therefore, Na, NO, and the excess Ag ions will all be present after precipitation of AgCl is complete. b. Ag (aq) Cl (aq) AgCl(s) c. Mass NaCl 0.61 g AgCl Mass % NaCl 0.61 g NaCl 1.50 g mixture 1 mol AgCl 1. g 1 mol Cl mol AgCl 100 17.% NaCl 1 mol NaCl mol Cl 58. g mol NaCl 0.61 g NaCl Acid-Base Reactions 71. All the bases in this problem are ionic compounds containing OH -. The acids are either strong or weak electrolytes. The best way to determine if an acid is a strong or weak electrolyte is to memorize all the strong electrolytes (strong acids). Any other acid you encounter that is not a strong acid will be a weak electrolyte (a weak acid), and the formula should be left unaltered in the complete ionic and net ionic equations. The strong acids to recognize are HCl, HBr, HI, HNO, HClO, and H SO. For the following answers, the order of the equations are formula, complete ionic, and net ionic. a. HClO (aq) Mg(OH) (s) H O(l) Mg(ClO ) (aq) H (aq) ClO (aq) Mg(OH) (s) H O(l) Mg (aq) ClO (aq) H (aq) Mg(OH) (s) H O(l) Mg (aq)
118 CHAPTER SOUTION STOICHIOMETRY b. HCN(aq) NaOH(aq) H O(l) NaCN(aq) HCN(aq) Na (aq) OH (aq) H O(l) Na (aq) CN (aq) HCN(aq) OH (aq) H O(l) CN (aq) c. HCl(aq) NaOH(aq) H O(l) NaCl(aq) H (aq) Cl (aq) Na (aq) OH (aq) H O(l) Na (aq) Cl (aq) H (aq) OH (aq) H O(l) 7. a. HNO (aq) Al(OH) (s) H O(l) Al(NO ) (aq) H (aq) NO (aq) Al(OH) (s) H O(l) Al (aq) NO (aq) H (aq) Al(OH) (s) H O(l) Al (aq) b. HC H O (aq) KOH(aq) H O(l) KC H O (aq) HC H O (aq) K (aq) OH (aq) H O(l) K (aq) C H O (aq) HC H O (aq) OH (aq) H O(l) C H O (aq) c. Ca(OH) (aq) HCl(aq) H O(l) CaCl (aq) Ca (aq) OH (aq) H (aq) Cl (aq) H O(l) Ca (aq) Cl (aq) H (aq) OH (aq) H O(l) or H (aq) OH (aq) H O(l) 7. All the acids in this problem are strong electrolytes (strong acids). The acids to recognize as strong electrolytes are HCl, HBr, HI, HNO, HClO, and H SO. a. KOH(aq) HNO (aq) H O(l) KNO (aq) b. Ba(OH) (aq) HCl(aq) H O(l) BaCl (aq) c. HClO (aq) Fe(OH) (s) H O(l) Fe(ClO ) (aq) d. AgOH(s) HBr(aq) AgBr(s) H O(l) e. Sr(OH) (aq) HI(aq) H O(l) SrI (aq) 7. a. Perchloric acid plus potassium hydroxide is a possibility. HClO (aq) KOH(aq) H O(l) KClO (aq) b. Nitric acid plus cesium hydroxide is a possibility. HNO (aq) CsOH(aq) H O(l) CsNO (aq)
CHAPTER SOUTION STOICHIOMETRY 119 c. Hydroiodic acid plus calcium hydroxide is a possibility. HI(aq) Ca(OH) (aq) H O(l) CaI (aq) 75. If we begin with 50.00 m of 0.00 M NaOH, then: 50.00 10 0.00 mol a. NaOH(aq) HCl(aq) NaCl(aq) H O(l) 1.00 10 mol NaOH b. HNO (aq) NaOH(aq) H O(l) NaNO (aq) 1.00 10 mol NaOH is to be neutralized 1 mol HCl 1 0.100 or 100. m mol NaOH 0.100 mol 1.00 10 mol NaOH 1 mol HNO mol NaOH 1 6.67 10 or 66.7 m 0.150 mol HNO c. HC H O (aq) NaOH(aq) H O(l) NaC H O (aq) 1.00 10 mol NaOH 1 mol HC H O mol NaOH 1 0.00 mol HC H O 5.00 10 50.0 m 76. We begin with 5.00 m of 0.00 M HCl or 5.00 10 0.00 mol/ 5.00 10 mol HCl. a. HCl(aq) NaOH(aq) H O(l) NaCl(aq) 5.00 10 mol HCl b. HCl(aq) Sr(OH) (aq) H O(l) SrCl (aq) 5.00 10 mol HCl 1 mol HCl 1 5.00 10 or 50.0 m mol NaOH 0.100 mol NaOH 1 mol Sr(OH) mol HCl c. HCl(aq) KOH(aq) H O(l) KCl(aq) 5.00 10 mol HCl 1 mol KOH mol HCl 1 0.0500 mol Sr(OH) 5.00 10 50.0 m 1.00 10 0.0 m 0.50 mol KOH 77. Ba(OH) (aq) HCl(aq) BaCl (aq) H O(l); H (aq) OH (aq) H O(l) 75.0 10 0.50 mol HCl 1.88 10 mol HCl 1.88 10 mol H 1.88 10 mol Cl 5.0 10 0.0550 mol Ba(OH) 1. 10 mol Ba(OH) 1. 10 mol Ba.8 10 mol OH
10 CHAPTER SOUTION STOICHIOMETRY The net ionic equation requires a 1 : 1 mole ratio between OH and H. The actual mole OH to mole H ratio is greater than 1 : 1, so OH is in excess. Because 1.88 10 mol OH will be neutralized by the H, we have (.8 1.88) 10 0.60 10 mol OH in excess. M OH mol OH excess total volume 6.0 10 mol OH.0 0.0750 0.50 10 M OH 78. HCl and HNO are strong acids; Ca(OH) and RbOH are strong bases. The net ionic equation that occurs is H (aq) OH (aq) H O(l). Mol H 0.100 mol HCl 0.0500 0.1000 Mol OH 0.5000 0.00 mol HNO 0.0100 mol Ca(OH) 0.100 mol RbOH 0.000 1 mol H mol HCl 1 mol H 0.00500 0.000 0.050 mol H mol HNO mol OH mol Ca(OH) 1 mol OH 0.0100 0.000 0.000 mol OH mol RbOH We have an excess of OH, so the solution is basic (not neutral). The moles of excess OH 0.000 mol OH initially 0.050 mol OH reacted (with H ) 0.0050 mol OH excess. M OH 0.0050 mol OH (0.0500 0.1000 0.5000 0.000) 79. HCl(aq) NaOH(aq) H O(l) NaCl(aq).16 10 NaOH Molarity of HCl.56 10 0.106 mol NaOH NaOH 5.00 10 mol 0.0050 mol 0.8500 5.9 10 M 1 mol HCl.56 10 mol HCl mol NaOH 0.10 M HCl 80. HC H O (aq) NaOH(aq) H O(l) NaC H O (aq) a. 16.58 10 soln Concentration of HC H O (aq) 0.506 mol NaOH 1 mol HC H O soln mol NaOH 8.9 10 mol 0.01000 b. If we have 1.000 of solution: Total mass 1000. m Mass of HC H O 0.89 mol Mass % acetic acid 50.0 g 1006 g 60.05 g mol 100 5.010% 8.9 10 mol HC H O 0.89 M 1.006 g m 50.0 g HC H O 1006 g solution
CHAPTER SOUTION STOICHIOMETRY 11 81. HNO (aq) Ca(OH) (aq) H O(l) Ca(NO ) (aq) 5.00 10 HNO 0.0500 mol HNO HNO 1 mol Ca(OH) mol HNO 1 Ca(OH) 0.000 mol Ca(OH) 0.08.8 m Ca(OH) 8. Strong bases contain the hydroxide ion (OH ). The reaction that occurs is H OH H O. 0.150 mol H 0.010 1 mol OH mol H 1.80 10 mol OH The 0.0 m of the unknown strong base contains 1.80 10 mol OH. 1.80 10 mol OH 0.000 0.0600 M OH The unknown base concentration is one-half the concentration of OH ions produced from the base, so the base must contain OH in each formula unit. The three soluble strong bases that have two OH ions in the formula are Ca(OH), Sr(OH), and Ba(OH). These are all possible identities for the strong base. 8. KHP is a monoprotic acid: NaOH(aq) KHP(aq) H O(l) NaKP(aq) Mass KHP 0.006 NaOH 0.1000 mol NaOH NaOH 1 mol KHP mol NaOH 0. g KHP mol KHP 0.178 g KHP 8. Because KHP is a monoprotic acid, the reaction is (KHP is an abbreviation for potassium hydrogen phthalate): NaOH(aq) KHP(aq) NaKP(aq) H O(l) 0.108 g KHP 1 mol KHP 0. g KHP 1 mol NaOH 5.98 10 mol NaOH mol KHP There are 5.98 10 mol of sodium hydroxide in.67 m of solution. Therefore, the concentration of sodium hydroxide is: 5.98 10.67 10 mol 1.58 10 M NaOH Oxidation-Reduction Reactions 85. Apply the rules in Table.. a. KMnO is composed of K and MnO ions. Assign oxygen an oxidation state of, which gives manganese a 7 oxidation state because the sum of oxidation states for all atoms in MnO must equal the 1 charge on MnO. K, 1; O, ; Mn, 7.
1 CHAPTER SOUTION STOICHIOMETRY b. Assign O a oxidation state, which gives nickel a oxidation state. Ni, ; O,. c. Na Fe(OH) 6 is composed of Na cations and Fe(OH) 6 anions. Fe(OH) 6 is composed of an iron cation and 6 OH anions. For an overall anion charge of, iron must have a oxidation state. As is usually the case in compounds, assign O a oxidation state and H a 1 oxidation state. Na, 1; Fe, ; O, ; H, 1. d. (NH ) HPO is made of NH cations and HPO anions. Assign 1 as the oxidation state of H and as the oxidation state of O. In NH, x (1) 1, x oxidation state of N. In HPO, 1 y (), y 5 oxidation state of P. e. O, ; P, f. O, ; x () 0, x 8/ oxidation state of Fe; this is the average oxidation state of the three iron ions in Fe O. In the actual formula unit, there are two Fe ions and one Fe ion. g. O, ; F, 1; Xe, 6 h. F, 1; S, i. O, ; C, j. H, 1; O, ; C, 0 86. a. UO : O, ; for U, x (), x 6 b. As O : O, ; for As, (x) () 0, x c. NaBiO : Na, 1; O, ; for Bi, 1 x () 0, x 5 d. As : As, 0 e. HAsO : Assign H 1 and O ; for As, 1 x () 0, x f. Mg P O 7: Composed of Mg ions and P O 7 ions. Mg, ; O, ; P, 5 g. Na S O : Composed of Na ions and S O ions. Na, 1; O, ; S, h. Hg Cl : Hg, 1; Cl, 1 i. Ca(NO ) : Composed of Ca ions and NO ions. Ca, ; O, ; N, 5 87. a. b. c. (x) (1) 0, x d. e. 1 f. g. h. 5 i. 0 88. a. SrCr O 7: Composed of Sr and Cr O 7 ions. Sr, ; O, ; Cr, x 7(), x 6 b. Cu, ; Cl, 1 c. O, 0; d. H, 1; O, 1 e. Mg and CO ions present. Mg, ; O, ; C, ; f. Ag, 0 g. Pb and SO ions present. Pb, ; O, ; S, ; h. O, ; Pb,
CHAPTER SOUTION STOICHIOMETRY 1 i. Na and C O ions present. Na, 1; O, ; C, x (), x j. O, ; C, k. Ammonium ion has a 1 charge (NH ), and sulfate ion has a charge (SO ). Therefore, the oxidation state of cerium must be (Ce ). H, 1; N, ; O, ; S, 6 l. O, ; Cr, 89. To determine if the reaction is an oxidation-reduction reaction, assign oxidation states. If the oxidation states change for some elements, then the reaction is a redox reaction. If the oxidation states do not change, then the reaction is not a redox reaction. In redox reactions, the species oxidized (called the reducing agent) shows an increase in oxidation states, and the species reduced (called the oxidizing agent) shows a decrease in oxidation states. Redox? Oxidizing Reducing Substance Substance Agent Agent Oxidized Reduced a. Yes Ag Cu Cu Ag b. No c. No d. Yes SiCl Mg Mg SiCl (Si) e. No In b, c, and e, no oxidation numbers change. 90. The species oxidized shows an increase in oxidation states and is called the reducing agent. The species reduced shows a decrease in oxidation states and is called the oxidizing agent. The pertinent oxidation states are listed by the substance oxidized and the substance reduced. Redox? Oxidizing Reducing Substance Substance Agent Agent Oxidized Reduced a. Yes H O CH CH (C, ) H O (H, 1 0) b Yes AgNO Cu Cu (0 ) AgNO (Ag, 1 0) c. Yes HCl Zn Zn (0 ) HCl (H, 1 0) d. No; there is no change in any of the oxidation numbers. 91. Use the method of half-reactions described in Section.10 of the text to balance these redox reactions. The first step always is to separate the reaction into the two half-reactions, and then to balance each half-reaction separately. a. I I e ClO Cl Adding the two balanced half-reactions so electrons cancel: I (aq) H (aq) ClO (aq) I (aq) Cl (aq) H O(l) e H ClO Cl H O
1 CHAPTER SOUTION STOICHIOMETRY b. As O H AsO NO NO H O As O H AsO H NO NO H O eft O; right 8 O ( e H NO NO H O) Right hand side has 5 extra O. Balance the oxygen atoms first using H O, then balance H using H, and finally, balance charge using electrons. This gives: (5 H O As O H AsO H e ) Common factor is a transfer of 1 e. Add half-reactions so that electrons cancel. 1 e 16 H NO NO 8 H O 15 H O As O 6 H AsO 1 H 1 e 7 H O(l) H (aq) As O (s) NO (aq) NO(g) 6 H AsO (aq) c. ( Br Br e ) 5 MnO Mn H O (5 e 8 H MnO Mn H O) Common factor is a transfer of 10 e. 10 Br 5 Br 10 e 10 e 16 H MnO Mn 8 H O 16 H (aq) MnO (aq) 10 Br (aq) 5 Br (l) Mn (aq) 8 H O(l) d. CH OH CH O Cr O 7 Cr (CH OH CH O H e ) 1 H Cr O 7 Cr 7 H O 6 e 1 H Cr O 7 Cr 7 H O Common factor is a transfer of 6 e. CH OH CH O 6 H 6 e 6 e 1 H Cr O 7 Cr 7 H O 8 H (aq) CH OH(aq) Cr O 7 (aq) Cr (aq) CH O(aq) 7 H O(l) 9. a. (Cu Cu e ) NO NO H O ( e H NO NO H O) Adding the two balanced half-reactions so that electrons cancel: Cu Cu 6 e 6 e 8 H NO NO H O Cu(s) 8 H (aq) NO (aq) Cu (aq) NO(g) H O(l) b. ( Cl Cl e ) Cr O 7 Cr 7 H O 6 e 1 H Cr O 7 Cr 7 H O Add the two half-reactions with six electrons transferred:
CHAPTER SOUTION STOICHIOMETRY 15 6 Cl Cl 6 e 6 e 1 H Cr O 7 Cr 7 H O 1 H (aq) Cr O 7 (aq) 6 Cl (aq) Cl (g) Cr (aq) 7 H O(l) c. Pb PbSO PbO PbSO Pb H SO PbSO H PbO H SO PbSO H O Pb H SO PbSO H e e H PbO H SO PbSO H O Add the two half-reactions with two electrons transferred: e H PbO H SO PbSO H O Pb H SO PbSO H e Pb(s) H SO (aq) PbO (s) PbSO (s) H O(l) This is the reaction that occurs in an automobile lead-storage battery. d. Mn MnO ( H O Mn MnO 8 H 5 e ) NaBiO Bi Na NaBiO Bi Na 6 H NaBiO Bi Na H O ( e 6 H NaBiO Bi Na H O) 5 8 H O Mn MnO 16 H 10 e 10 e 0 H 5 NaBiO 5 Bi 5 Na 15 H O 8 H O 0 H Mn 5 NaBiO MnO 5 Bi 5 Na 15 H O 16 H Simplifying: 1 H (aq) Mn (aq) 5 NaBiO (s) MnO (aq) 5 Bi (aq) 5 Na (aq) 7 H O(l) e. H AsO AsH (Zn Zn e ) H AsO AsH H O 8 e 8 H H AsO AsH H O 8 e 8 H H AsO AsH H O Zn Zn 8 e 8 H (aq) H AsO (aq) Zn(s) Zn (aq) AsH (g) H O(l) 9. Use the same method as with acidic solutions. After the final balanced equation, convert H to OH as described in Section.10 of the text. The extra step involves converting H into H O by adding equal moles of OH to each side of the reaction. This converts the reaction to a basic solution while still keeping it balanced. a. Al Al(OH) MnO MnO H O Al Al(OH) H e H MnO MnO H O H O Al Al(OH) H e
16 CHAPTER SOUTION STOICHIOMETRY H O Al Al(OH) H e e H MnO MnO H O H O(l) Al(s) MnO (aq) Al(OH) (aq) MnO (s) H doesn t appear in the final balanced reaction, so we are done. b. Cl Cl Cl OCl e Cl Cl H O Cl OCl H e e Cl Cl H O Cl OCl H e H O Cl Cl OCl H Now convert to a basic solution. Add OH to both sides of the equation. The OH will react with the H on the product side to give H O. After this step, cancel identical species on both sides ( H O). Applying these steps gives: OH Cl Cl OCl H O, which can be further simplified to: OH (aq) Cl (g) Cl (aq) OCl (aq) H O(l) c. NO NH Al AlO 6 e 7 H NO NH H O ( H O Al AlO H e ) Common factor is a transfer of 6 e. 6e 7 H NO NH H O H O Al AlO 8 H 6 e OH H O NO Al NH AlO H OH Reducing gives OH (aq) H O(l) NO (aq) Al(s) NH (g) AlO (aq). 9. a. Cr Cr(OH) CrO Cr(OH) H O Cr Cr(OH) H e e 5 H CrO Cr(OH) H O H O Cr Cr(OH) H e e 5 H CrO Cr(OH) H O OH H H O Cr CrO Cr(OH) OH Two OH were added above to each side to convert to a basic solution. The two OH react with the H on the reactant side to produce H O. The overall balanced equation is: H O(l) Cr(s) CrO (aq) Cr(OH) (s) OH (aq) b. S S MnO MnS (S S e ) 5 MnO S MnS ( 5 e 8 H MnO S MnS H O) Common factor is a transfer of 10 e.
CHAPTER SOUTION STOICHIOMETRY 17 5 S 5 S 10 e 10 e 16 H MnO S MnS 8 H O 16 OH 16 H 7 S MnO 5 S MnS 8 H O 16 OH 16 H O 7 S MnO 5 S MnS 8 H O 16 OH Reducing gives 8 H O(l) 7 S (aq) MnO (aq) 5 S(s) MnS(s) 16 OH (aq). c. CN CNO (H O CN CNO H e ) MnO MnO ( e H MnO MnO H O) Common factor is a transfer of 6 electrons. H O CN CNO 6 H 6 e 6 e 8 H MnO MnO H O OH H CN MnO CNO MnO H O OH Reducing gives: H O(l) CN (aq) MnO (aq) CNO (aq) MnO (s) OH (aq) 95. NaCl H SO MnO Na SO MnCl Cl H O We could balance this reaction by the half-reaction method, which is generally the preferred method. However, sometimes a redox reaction is not so complicated and thus balancing by inspection is a possibility. et s try inspection here. To balance Cl, we need NaCl: NaCl H SO MnO Na SO MnCl Cl H O Balance the Na and SO ions next: NaCl H SO MnO Na SO MnCl Cl H O On the left side: H and 10 O; on the right side: 8 O not counting H O We need H O on the right side to balance H and O: NaCl(aq) H SO (aq) MnO (s) Na SO (aq) MnCl (aq) Cl (g) H O(l) 96. Au HNO HCl AuCl NO Only deal with ions that are reacting (omit H ): Au NO Cl AuCl NO The balanced half-reactions are: Au Cl AuCl e e H NO NO H O Adding the two balanced half-reactions: Au(s) Cl (aq) H (aq) NO (aq) AuCl (aq) NO(g) H O(l)
18 CHAPTER SOUTION STOICHIOMETRY 97. (H C O CO H e ) 5 (5 e 8 H MnO Mn H O) 5 H C O 10 CO 10 H 10 e 10 e 16 H MnO Mn 8 H O 6 H (aq) 5 H C O (aq) MnO (aq) 10 CO (g) Mn (aq) 8 H O(l) 0.1058 g H C O Molarity 1 mol H C 90.0 g O.700 10 mol MnO 8.97 m mol MnO.700 10 mol MnO 5 mol H C O 1000 m 98. (Fe Fe e ) 5 5 e 8 H MnO Mn H O 1.6 10 M MnO 8 H (aq) MnO (aq) 5 Fe (aq) 5 Fe (aq) Mn (aq) H O(l) From the titration data we can get the number of moles of Fe. We then convert this to a mass of iron and calculate the mass percent of iron in the sample. 8.7 10 MnO 0.0198 mol MnO.80 10 55.85 g Fe mol Fe 0.1 g Fe mol Fe Mass % Fe 0.1 g 0.618 g 100.6% Fe 5 mol Fe mol MnO.80 10 mol Fe.80 10 mol Fe present 99. a. (Fe Fe e ) 5 5 e 8 H MnO Mn H O The balanced equation is: 8 H (aq) MnO (aq) 5 Fe (aq) 5 Fe (aq) Mn (aq) H O(l) 0.6 10 soln Molarity. 10 50.00 10 0.016 mol MnO mol Fe soln 5 mol Fe mol MnO.6 10 M Fe. 10 mol Fe b. (Fe Fe e ) 6 6 e 1 H Cr O 7 Cr 7 H O The balanced equation is: 1 H (aq) Cr O 7 (aq) 6 Fe (aq) 6 Fe (aq) Cr (aq) 7 H O(l) 50.00 10.6 10 mol Fe 1 mol CrO 7 1 6 mol Fe 0.0150 mol Cr O.8 10 or.8 m 7
CHAPTER SOUTION STOICHIOMETRY 19 100. (6 e 1 H Cr O 7 Cr 7 H O) H O C H 5OH CO 1 H 1 e 16 H (aq) Cr O 7 (aq) C H 5OH(aq) Cr (aq) CO (g) 11 H O(l) 0.0105 0.0600 mol Cr O7 1 mol CH5OH 6.07 g 0.09 g C H 5OH mol Cr O mol CH5OH 7 0 5.09 g C H OH 100 0.1% C H 5OH 0.0 g blood Additional Exercises 101. Desired uncertainty is 1% of 0.0, or ±0.000. So we want the solution to be 0.000 ± 0.000 M, or the concentration should be between 0.0198 and 0.00 M. We should use a 1- volumetric flask to make the solution. They are good to ±0.1%. We want to weigh out between 0.0198 mol and 0.00 mol of KIO. Molar mass of KIO 9.10 16.9 (16.00) 1.0 g/mol 0.0198 mol 1.0 g mol.7 g; 0.00 mol 1.0 g mol. g (carrying extra sig. figs.) We should weigh out between. and. g of KIO. We should weigh it to the nearest milligram, or nearest 0.1 mg. Dissolve the KIO in water, and dilute (with mixing along the way) to the mark in a 1- volumetric flask. This will produce a solution whose concentration is within the limits and is known to at least the fourth decimal place. 10. Solution A: molecules 1.0 ; solution B: 6 molecules.0 1.5 molecules 1.0 Solution C: molecules.0 molecules ; solution D: 1.0 6 molecules.0 molecules 1.0 Solution A has the most molecules per unit volume so solution A is most concentrated. This is followed by solution D, then solution C. Solution B has the fewest molecules per unit volume, so solution B is least concentrated. 10..0 g C 1H O 11 1 11 mol C1 HO 0.095 mol C 1H O 11 added to blood.0 g The blood sugar level would increase by: 0.095 mol C1 H O11 0.019 mol/ 5.0 0.75 mol CaCl 10. Mol CaCl present 0.0 CaCl CaCl 6. 10 mol CaCl
10 CHAPTER SOUTION STOICHIOMETRY The volume of CaCl solution after evaporation is: 6. 10 1 CaCl mol CaCl 1.10 mol CaCl 5.75 10 57.5 m CaCl Volume H O evaporated 0. m 57.5 m 17 m H O evaporated 105. CaCl (aq) Na C O (aq) CaC O (s) NaCl(aq) is the balanced formula equation. Ca (aq) C O (aq) CaC O (s) is the balanced net ionic equation. 106. CaCO (s) HCl(aq) CaCl (aq) H O(l) CO (g) 107. There are other possible correct choices for most of the following answers. We have listed only three possible reactants in each case. a. AgNO, Pb(NO ), and Hg (NO ) would form precipitates with the Cl ion. Ag (aq) Cl (aq) AgCl(s); Pb (aq) Cl (aq) PbCl (s) Hg (aq) Cl (aq) Hg Cl (s) b. Na SO, Na CO, and Na PO would form precipitates with the Ca ion. Ca (aq) SO (aq) CaSO (s); Ca (aq) CO (aq) CaCO (s) Ca (aq) PO (aq) Ca (PO ) (s) c. NaOH, Na S, and Na CO would form precipitates with the Fe ion. Fe (aq) OH (aq) Fe(OH) (s); Fe (aq) S (aq) Fe S (s) Fe (aq) CO (aq) Fe (CO ) (s) d. BaCl, Pb(NO ), and Ca(NO ) would form precipitates with the SO ion. Ba (aq) SO (aq) BaSO (s); Pb (aq) SO (aq) PbSO (s) Ca (aq) SO (aq) CaSO (s) e. Na SO, NaCl, and NaI would form precipitates with the Hg ion. Hg (aq) SO (aq) Hg SO (s); Hg (aq) Cl (aq) Hg Cl (s) Hg (aq) I (aq) Hg I (s) f. NaBr, Na CrO, and Na PO would form precipitates with the Ag ion. Ag (aq) Br - (aq) AgBr(s); Ag (aq) CrO (aq) Ag CrO (s) Ag (aq) PO (aq) Ag PO (s) 108. a. MgCl (aq) AgNO (aq) AgCl(s) Mg(NO ) (aq)
CHAPTER SOUTION STOICHIOMETRY 11 0.61 g AgCl 1 mol AgCl 1. g AgCl 0.1 g MgCl 100 1.% MgCl 1.50 g mixture 1 mol MgCl b. 0.1 g MgCl 95.1 g 1 mol MgCl 95.1 g 0.1 g MgCl mol AgCl mol MgCl mol AgNO mol MgCl 1 0.500 mol AgNO 109. XCl (aq) AgNO (aq) AgCl(s) X(NO ) (aq) 1.8 g AgCl.91 10 1.00 g XCl 1 mol AgCl 1. g mol XCl 1 mol XCl.81 10 mol XCl mol AgCl 08 g/mol; x (5.5) 08, x 17 g/mol From the periodic table, the metal X is barium (Ba). 1000 m 1 8.95 m AgNO 110. From the periodic table, use aluminum in the formulas to convert from mass of Al(OH) to mass of Al (SO ) in the mixture. 1 mol Al(OH) 0.107 g Al(OH) 78.00 g Mass % Al (SO ) 0.5 g 1.5 g 1 mol Al 1 mol Al (SO ) mol Al(OH) mol Al.17 g Al (SO ) 0.5 g Al (SO ) mol Al (SO ) 100 16.% 111. All the Tl in TlI came from Tl in Tl SO. The conversion from TlI to Tl SO uses the molar masses and formulas of each compound. 0. g Tl 50.9 g TlSO 0.18 g TlI 0.190 g Tl SO 1. g TlI 08.8 g Tl Mass % Tl SO 0.190 g Tl SO 100 1.65% Tl SO 9.86 g pesticide 11. a. Fe (aq) OH (aq) Fe(OH) (s) Fe(OH) : 55.85 (16.00) (1.008) 106.87 g/mol 55.85 g Fe 0.107 g Fe(OH) 0.0559 g Fe 106.87 g Fe(OH) b. Fe(NO ) : 55.85 (1.01) 9(16.00) 1.86 g/mol
1 CHAPTER SOUTION STOICHIOMETRY 0.0559 g Fe 1.86 g Fe(NO ) 0. g Fe(NO ) 55.85 g Fe c. Mass % Fe(NO ) 0. g 0.56 g 100 5.1% 11. With the ions present, the only possible precipitate is Cr(OH). Cr(NO ) (aq) NaOH(aq) Cr(OH) (s) NaNO (aq) Mol NaOH used.06 g Cr(OH) to form precipitate 1 mol Cr(OH) 10.0 g mol NaOH 6.00 10 mol mol Cr(OH) NaOH(aq) HCl(aq) NaCl(aq) H O(l) Mol NaOH used 0.1000 to react with HCl total mol NaOH M NaOH volume 0.00 mol HCl 6.00 10 11. a. MgO(s) HCl(aq) MgCl (aq) H O(l) 1 mol NaOH.00 10 mol mol HCl mol.00 10 0.0500 Mg(OH) (s) HCl(aq) MgCl (aq) H O(l) Al(OH) (s) HCl(aq) AlCl (aq) H O(l) mol.00 M NaOH b. et's calculate the number of moles of HCl neutralized per gram of substance. We can get these directly from the balanced equations and the molar masses of the substances. mol HCl mol MgO mol HCl mol Mg(OH) mol HCl mol Al(OH) 1 mol MgO 0.1g MgO.96 10 mol HCl g MgO 1 mol Mg(OH).9 10 58. g Mg(OH) g Mg(OH) 1 mol Al(OH).86 10 78.00 g Al(OH) g Al(OH) mol HCl mol HCl Therefore, 1 gram of magnesium oxide would neutralize the most 0.10 M HCl. 115. Using HA as an abbreviation for the monoprotic acid acetylsalicylic acid: HA(aq) NaOH(aq) H O(l) NaA(aq) 0.5065 mol NaOH 1 mol HA Mol HA 0.0517 NaOH 1.781 10 mol HA NaOH mol NaOH
CHAPTER SOUTION STOICHIOMETRY 1 From the problem,.10 g HA was reacted, so: molar mass.10 g HA 1.781 10 mol HA 180. g/mol 116. Mg(s) HCl(aq) MgCl (aq) H (g).00 g Mg 1 mol Mg.1g Mg mol HCl 1 0.09 9 m HCl mol Mg 5.0 mol HCl 117. et HA unknown monoprotic acid; HA(aq) NaOH(aq) NaA(aq) H O(l) 0.500 mol NaOH Mol HA present 0.050 1 mol HA 0.015 mol HA 1 mol NaOH x g HA mol HA.0 g HA 0.015 mol HA, x molar mass of HA 176 g/mol Empirical formula weight (1) (1) (16) 88 g/mol. Because 176/88.0, the molecular formula is (C H O ) C 6H 8O 6. 118. We get the empirical formula from the elemental analysis. Out of 100.00 g carminic acid, there are: 5.66 g C 1 mol C 1.01g C.68 mol C;.09 g H 1 mol H 1.008 g H.06 mol H.5 g O 1 mol O 16.00 g O.61 mol O Dividing the moles by the smallest number gives:.68.61 1.69;.06.61 1.5 These numbers don t give obvious mole ratios. et s determine the mol C to mol H ratio:.68.06 1.10. 06 So let's try 10 11 10.68.06.61 0.06 as a common factor: 11.0; 10.0; 6. 50 0.06 0.06 0.06 Therefore, C H 0O 1 is the empirical formula. We can get molar mass from the titration data. The balanced reaction is HA(aq) OH (aq) H O(l) A (aq), where HA is an abbreviation for carminic acid, an acid with one acidic proton (H ).
1 CHAPTER SOUTION STOICHIOMETRY 18.0 10 soln Molar mass 0.006 mol NaOH soln 0.60 g 7. 10 mol 9 g mol 1 mol carminic acid mol NaOH 7. 10 mol carminic acid The empirical formula mass of C H 0O 1 (1) 0(1) 1(16) 9 g. Therefore, the molecular formula of carminic acid is also C H 0O 1. 119. 0.10 g AgCl 1mol AgCl 1mol Cl 5.5 g Cl.57 10 g Cl 1. g AgCl mol AgCl mol Cl All of the Cl in the AgCl precipitate came from the chlorisondamine chloride compound in the medication. So we need to calculate the quantity of C 1H 0Cl 6N which contains.57 10 g Cl. Molar mass of C 1H 0Cl 6N 1(1.01) 0(1.008) 6(5.5) (1.01) 9.0 g/mol There are 6(5.5) 1.70 g chlorine for every mole (9.0 g) of C 1H 0Cl 6N..57 10 g Cl Mass % chlorisondamine chloride 9.0 g C1 H 0Cl6N 5.18 10 g C 1H 0Cl 6N 1.70 g Cl 5.18 10 1.8 g g 100.05% 10. All the sulfur in BaSO came from the saccharin. The conversion from BaSO to saccharin utilizes the molar masses of each compound. 0.50 g BaSO.07 g S 18.19 g C7H5NOS 0.99 g C 7H 5NO S. g BaSO.07 g S Average mass Tablet 0.99 g 10 tablets.99 10 tablet g 9.9 mg tablet Average mass %.99 g C7 H5NO S 100 67.00% saccharin by mass 0.589 g 0 11. Use the silver nitrate data to calculate the mol Cl present, then use the formula of douglasite (KCl FeCl H O) to convert from Cl to douglasite (1 mole of douglasite contains moles of Cl ). The net ionic equation is Ag Cl AgCl(s). 0.070 Mass % douglasite 0.1000 mol Ag 1 mol Cl mol Ag 0.900 g 100 6.7% 0.550 g 1 mol douglasite mol Cl 11.88 g douglasite mol 0.900 g douglasite