CHAPTER TYPES OF CHEMICA REACTIONS AND SOUTION STOICHIOMETRY Questions 1. 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 ecule. Polar ecules 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 ecule and dissolves other polar solutes readily. The oxygen end of water (the partial negative end of the polar water ecule) aligns with the partial positive end of the polar solute, whereas the hydrogens of water (the partial positive end of the polar water ecule) 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 6 H 1 O 6 is a polar covalent ecule that is a nonelectrolyte. C 6 H 1 O 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 ecule and is hydrated as explained in part a. 1..0.0 / = 6.0 HCl; the.0 of solution contains 6.0 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 of H + (aq) and 6.0 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 es of HC H O ecules 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 ecules but will also contain some hydrated H + ions and hydrated C H O ions. 15. 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. 9
9 CHAPTER SOUTION STOICHIOMETRY 16. One e of NaOH dissolved in 1.00 of solution will produce 1.00 M NaOH. First, weigh out 0.00 g of NaOH (1.000 ). 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 arity will be known to four significant figures. This is good practice, if you need a three-significant-figure arity, 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 arity. Here, we need to cut the arity in half from.00 M to 1.00 M. We would start with 1 e 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. 17. 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. 18. 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 of Pb + ions would react with the.0 of I ions to form 1.0 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. 19. 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. 0. 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.
CHAPTER SOUTION STOICHIOMETRY 95 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). 1. 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.. Reference the Problem Solving Strategy box in Section.10 of the text for the steps involved in balancing redox reactions by oxidation states. The key to the oxidation states method is to balance the electrons gained by the species reduced with the number of electrons lost from the species oxidized. This is done by assigning oxidation states and, from the change in oxidation states, determining the coefficients necessary to balance electrons gained with electrons lost. After the loss and gain of electrons is balanced, the remainder of the equation is balanced by inspection. Exercises Aqueous Solutions: Strong and Weak Electrolytes. 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.
96 CHAPTER SOUTION STOICHIOMETRY 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 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). 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. 5. CaCl (s) Ca + (aq) + Cl (aq) 6. MgSO (s) Mg + (aq) + SO (aq); NH NO (s) NH + (aq) + NO (aq) Solution Concentration: Molarity 7. a. 5.6 g NaHCO 1 NaHCO 8.01g NaHCO = 6.69 10 NaHCO M = 6.69 10 50.0 m 1000m = 0.677 M NaHCO
CHAPTER SOUTION STOICHIOMETRY 97 b. 0.186 g K Cr O 7 1K 9.0g K Cr Cr O 7 O 7 = 6.75 10 K Cr O 7 M = 6.75 10 500.0 10 = 1.55 10 M K Cr O 7 c. 0.105 g Cu 1 Cu 6.55g Cu = 1.61 10 Cu = 1.61 10 Cu + M = 1.61 10 Cu 00.0 m 1000m = 8.065 10 M Cu + 8. 75.0 m 0.79g m 1 = 1. C H 5 OH; arity = 6.07g 1. 0.50 = 5. M C H 5 OH 9. a. M Ca(NO ) = 0.100Ca(NO ) 0.100 = 1.00 M Ca(NO ) (s) Ca + (aq) + NO (aq); M = 1.00 M; M = (1.00) =.00 M Ca NO b. M Na SO =.5 Na SO 1.5 =.0 M Na SO (s) Na + (aq) + SO (aq); M = (.0) =.0 M ; M =.0 M Na SO c. 5.00 g NH Cl 1 NH 5.9g NH Cl Cl = 0.095 NH Cl M NH Cl = 0.095 NH Cl 0.5000 = 0.187 M NH Cl(s) NH + (aq) + Cl (aq); M = M = 0.187 M NH Cl d. 1.00 g K PO 1K PO 1.7g =.71 10 K PO M K PO =.71 10 0.500 = 0.0188 M K PO (s) K + (aq) + PO (aq); M = (0.0188) = 0.056 M; M = 0.0188 M K PO
98 CHAPTER SOUTION STOICHIOMETRY 0. a. M Na PO = 0.000 =.00 M 0.0100 PO Na PO (s) Na + (aq) + (aq) M = (.00) = 6.00 M; M =.00 M ; Na PO b. M Ba(NO ) = 0.00 0.6000 = 0.500 M Ba(NO ) (s) Ba + (aq) + NO (aq); M = 0.500 M; Ba M = (0.500) = 1.00 M NO c. M KCl = 1.00g KCl 0.5000 1 KCl 7.55g KCl = 0.068 M KCl(s) K + (aq) + Cl (aq); M M = 0.068 M K Cl d. M = (NH ) SO 1 (NH) SO 1g (NH) SO 1.15g 1.50 = 0.666 M (NH ) SO (s) NH + (aq) + SO (aq) M = (0.666) = 1. M; M = 0.666 M NH SO 1. Mol solute = volume () arity ; AlCl (s) Al + (aq) + Cl (aq) Mol Cl = 0.1000 0.0 AlCl Cl = 9.0 10 Cl AlCl MgCl (s) Mg + (aq) + Cl (aq) Mol Cl = 0.0500 0.60 MgCl Cl = 6.0 10 Cl MgCl NaCl(s) Na + (aq) + Cl (aq) Mol Cl = 0.000 0.0 NaCl 1Cl = 8.0 NaCl 100.0 m of 0.0 M AlCl contains the most es of Cl ions. 10 Cl. NaOH(s) Na + (aq) + OH (aq), total of ions (1 Na + and 1 Cl ) per NaOH.
CHAPTER SOUTION STOICHIOMETRY 99 0.1000 0.100 NaOH ions =.0 NaOH 10 ions BaCl (s) Ba + (aq) + Cl (aq), total of ions per BaCl. 0.0500 0.00 ions =.0 BaCl 10 ions Na PO (s) Na + (aq) + PO (aq), total of ions per Na PO. 0.0750 0.150 NaPO ions =.50 10 ions Na PO 75.0 m of 0.150 M Na PO contains the largest number of ions.. Molar mass of NaOH =.99 + 16.00 + 1.008 = 0.00 g/ Mass NaOH = 0.500 0.00 NaOH 0.00g NaOH =.00 g NaOH NaOH. 10. g AgNO 1AgNO 169.9 g 1 = 0. = 0 m 0.5AgNO 5. a..00 0.50 NaOH 0.00g NaOH = 0.0 g NaOH 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 NaOH 1 stock = 0.500 1.00 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. c..00 0.100 KCrO 19.0g KCrO = 8.8 g K CrO K CrO Similar to the solution made in part a, instead using 8.8 g K CrO. d..00 0.100 KCrO 1 stock = 0.11 1.75 K CrO Similar to the solution made in part b, instead using 11 m of the 1.75 M K CrO stock solution.
100 CHAPTER SOUTION STOICHIOMETRY 6. a. 1.00 solution 0.50 H SO 0.50HSO 1 18H SO = 0.50 H SO =.8 10 conc. H SO or 8 m 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 HCl. 0.50 HCl 1 1 HCl =. 10 = m Dilute m of concentrated HCl to a final volume of 1.00. c. We need 0.50 NiCl. 1 NiCl 6HO 0.50 NiCl NiCl 7.69g NiCl NiCl 6HO 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. d. 1.00 0.50 HNO = 0.50 HNO 0.50 HNO 1 16 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 Na CO. 0.50 Na CO 105.99g Na CO = 5 g Na CO Dissolve 5 g Na CO in water, dilute to 1.00. 7. 10.8 g (NH ) SO 1 1.15g = 8.17 10 (NH ) SO Molarity = 8.1710 1000m = 0.817 M (NH ) SO 100.0 m
CHAPTER SOUTION STOICHIOMETRY 101 Moles of (NH ) SO in final solution: 10.00 10-0.817 Molarity of final solution = = 8.17 (NH ) SO (s) NH + (aq) + SO (aq); 10 8.17 10 1000m (10.00 50.00) m = 0.16 M (NH ) SO M = (0.16) = 0.7 M; M = 0.16 M total HNO 8. Molarity = ; total volume = 0.05000 + 0.10000 = 0.15000 totalvolume NH 0.100 HNO Total HNO = 0.05000 + 0.10000 SO 0.00 HNO Total HNO = 5.00 10 +.00 10 =.50 10 HNO Molarity =.5010 HNO 0.15000 = 0.167 M HNO As expected, the arity of HNO is between 0.100 M and 0.00 M. 9. Mol Na CO = 0.0700.0 Na CO = 0.1 Na CO Na CO (s) Na + (aq) + CO (aq); Na + = (0.1 ) = 0. 1.0 NaHCO Mol NaHCO = 0.000 = 0.00 NaHCO NaHCO (s) Na + (aq) + HCO (aq); Na + = 0.00 M = Na total Na totalvolume = 0. 0.00 0.0700 0.000 0.5 0.1000 =.5 M Na + 0. Mol CoCl = 0.0500 0.50CoCl = 0.015 Mol NiCl = 0.050 0.50 NiCl = 0.00875 Both CoCl and NiCl are soluble chloride salts by the solubility rules. A 0.015- aqueous sample of CoCl is actually 0.015 Co + and (0.015 ) = 0.050 Cl. A 0.00875- aqueous sample of NiCl is actually 0.00875 Ni + and (0.00875) = 0.0175 Cl. The total volume of solution that these ions are in is 0.0500 + 0.050 = 0.0750.
10 CHAPTER SOUTION STOICHIOMETRY 0.015Co 0.00875 Ni M 0.167 M; M 0. 117M Co 0.0750 Ni 0.0750 0.050Cl 0.0175Cl M 0. 567 M Cl 0.0750 1. Stock solution = 10.0 mg 500.0 m 10.0 10 g 500.0 m 5.00 10 g steroid m 100.0 6 10 stock 1000m This is diluted to a final volume of 100.0 m. 5.00 10 g steroid =.00 m 6 10 g steroid 6.00 10 g steroid 100.0 m 1000m 1 steroid 6.g steroid = 5.9 8 10 M steroid. Stock solution: 1.58 g Mn + 1 Mn 5.9g Mn =.88 10 Mn + Molarity = Solution A:.8 10 Mn 1.000 =.88 10 M 50.00 m 1 1000m.8 10 = 1. 10 Mn + Molarity = Solution B: 1. 10 1000.0 m 1000m 1 = 1. 10 M 10.0 m 1 1000m 1. 10 5 = 1. 10 Mn + Molarity = Solution C: 5 1. 10 0.500 = 5.768 5 10 M 10.00 10-5.768 10 5 = 5.768 7 10 Mn + Molarity = 7 5.768 10 0.5000 = 1.15 6 10 M
CHAPTER SOUTION STOICHIOMETRY 10 Precipitation Reactions. 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. 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).. 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. Inoluble (Rule ) d. Soluble (Rules and ) e. Insoluble (Rule 6) f. Insoluble (Rule 5) g. Insoluble (Rule 6) h. Soluble (Rule ) 5. 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) 6. 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.
10 CHAPTER SOUTION STOICHIOMETRY c. Possible products = KI and MgCO ; precipitate = MgCO d. Possible products = NaBr and Al (CrO ) ; precipitate = Al (CrO ) 7. 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) 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) 8. 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) 9. 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)
CHAPTER SOUTION STOICHIOMETRY 105 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) 50. 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) 51. 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) 5. 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)
106 CHAPTER SOUTION STOICHIOMETRY Hg + (aq) + SO (aq) Hg SO (s) d. No reaction occurs because all possible products (SrI and KNO ) are soluble. 5. 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). + 5. 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)]. 55. AgNO (aq) + Na CrO (aq) Ag CrO (s) + NaNO (aq) 0.0750 0.100 AgNO 1 NaCrO 161.98g NaCrO = 0.607 g Na CrO AgNO Na CrO 56. Na PO (aq) + Pb(NO ) (aq) Pb (PO ) (s) + 6 NaNO (aq) 0.1500 0.50 Pb(NO) NaPO 1 NaPO = 0.50 Pb(NO ) 0.100 Na PO 57. A1(NO ) (aq) + KOH(aq) Al(OH) (s) + KNO (aq) = 50. m Na PO Assuming Al(NO ) is limiting: 0.0500 0.00 Al(NO ) Assuming KOH is limiting: 0.000 0.100 KOH 1Al(OH) Al(NO ) 1Al(OH) KOH 78.00g Al(OH) Al(OH) = 0.780 g Al(OH) 78.00g Al(OH) = 0.50 g Al(OH) Al(OH) Because KOH produces the smaller mass of the Al(OH) precipitate, KOH is the limiting reagent and 0.50 g Al(OH) can form. 58. The balanced equation is BaCl (aq) + Fe (SO ) (aq) BaSO (s) + FeCl (aq). 100.0 m BaCl 1 1000m 0.100 BaCl BaSO BaCl. g BaSO BaSO =. g BaSO
CHAPTER SOUTION STOICHIOMETRY 107 100.0 m Fe (SO ) 1000m 0.100 Fe(SO 1 ) BaSO Fe (SO ). g BaSO = 7.00 g BaSO BaSO The BaCl reagent produces the smaller quantity of the BaSO precipitate, so BaCl is limiting and. g BaSO can form. 59. The reaction is AgNO (aq) + NaBr(aq) AgBr(s) + NaNO (aq). Assuming AgNO is limiting: 100.0 m AgNO 1 1000m 0.150 AgNO AgNO 1AgBr AgNO 187.8 g AgBr AgBr =.8 g AgBr Assuming NaBr is limiting: 0.0 m NaBr 1 1000m 1.00 NaBr NaBr 1AgBr NaBr 187.8 g AgBr AgBr =.76 g AgBr The AgNO reagent produces the smaller quantity of AgBr, so AgNO is limiting and.8 g AgBr can form. 60. AgNO (aq) + CaCl (aq) AgCl(s) + Ca(NO ) (aq) 0.1000 0.0 AgNO AgCl 1. g AgCl =.9 g AgCl AgNO AgCl 0.1000 0.15CaCl AgCl CaCl 1. g AgCl AgCl =. g 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 es of each ion present initially (before reaction) can be easily determined from the es of each reactant. We have 0.1000 (0.0 AgNO /) = 0.00 AgNO, which dissolves to form 0.00 Ag + and 0.00 NO. We also have 0.1000 (0.15 CaCl /) = 0.015 CaCl, which dissolves to form 0.015 Ca + and (0.015) = 0.00 Cl. To form the.9 g of AgCl precipitate, 0.00 Ag + will react with 0.00 of Cl to form 0.00 AgCl (which has a mass of.9 g). Mol unreacted Cl = 0.00 Cl initially 0.00 Cl reacted Mol unreacted Cl = 0.010 Cl
108 CHAPTER SOUTION STOICHIOMETRY M = Cl 0.010Cl 0.010Cl = 0.050 M Cl totalvolume 0.1000 0.1000 The arities of the spectator ions are: 0.00 NO 0.000 = 0.10 M NO ; 0.015Ca 0.000 = 0.075 M Ca + 61. a. The balanced reaction is KOH(aq) + Mg(NO ) (aq) Mg(OH) (s) + KNO (aq). b. The precipitate is magnesium hydroxide. c. Assuming KOH is limiting: 0.1000 KOH 0.00 KOH KOH 1Mg(OH) KOH 58.g Mg(OH) Mg(OH) = 0.58 g Mg(OH) Assuming Mg(NO ) is limiting: 0.1000 Mg(NO ) 0.00 Mg(NO Mg(NO ) ) 1Mg(OH) Mg(NO ) 58.g Mg(OH) = 1.17 g Mg(OH) 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 Mg(NO) Mg(NO ) 1Mg Mg(NO ) = 0.000 Mg + Mol Mg + reacted = 0.1000 KOH 0.00 KOH 1Mg(NO ) KOH KOH 1Mg = 0.0100 Mg + Mg(NO ) M = Mg excess Mg totalvolume = (0.000 0.0100) Mg 0.1000 0.1000 = 5.00 10 M Mg +
CHAPTER SOUTION STOICHIOMETRY 109 The spectator ions are K + and NO. The es of each are: K + = 0.1000 KOH 0.00 KOH 1K = 0.000 K + KOH KOH NO = 0.1000 Mg(NO ) 0.00 Mg(NO) Mg(NO ) NO Mg(NO ) The concentrations are: = 0.000 NO 0.000 K 0.000 = 0.100 M K + ; 0.000NO 0.000 = 0.00 M NO 6. a. Cu(NO ) (aq) + KOH(aq) Cu(OH) (s) + KNO (aq) Solution A contains.00.00 / =.00 Cu(NO ), and solution B contains.00.00 / = 6.00 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 ecules 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 es, M = 6.00 K = 1.50 M and M = NO.00 8.00 NO.00 K =.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
110 CHAPTER SOUTION STOICHIOMETRY volume of solution doubled, the concentration of Cu + ions will decrease by a factor of eight after the two beakers are mixed: 1 M =.00 M Cu = 0.50 M 8 Alternately, one could certainly use es to solve for M : Cu Mol Cu + reacted =.00.00OH 1Cu OH =.00 Cu + reacted Mol Cu + present initially =.00.00Cu =.00 Cu + present initially Excess Cu + present after reaction =.00.00 = 1.00 Cu + excess M = Cu 1.00Cu.00.00 = 0.50 M Mass of precipitate = 6.00 KOH 6. M SO (aq) + CaCl (aq) CaSO (s) + MCl(aq) 1Cu(OH) KOH 97.57g Cu(OH) Cu(OH) = 9 g Cu(OH) 1.6 g CaSO 1CaSO 16.15g CaSO 1MSO = 9.99 10 M SO CaSO From the problem, 1. g M SO was reacted, so: ar mass = 9.99 10 1.g M SO M SO = 1 g/ 1 u = (atomic mass M) +.07 + (16.00), atomic mass M = u From periodic table, M = Na (sodium). 6. 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 1AgCl 1. g 1Cl AgCl 1 NaCl Cl 58.g NaCl = 0.61 g NaCl Mass % NaCl = 0.61g NaCl 1.50g mixture 100 = 17.% NaCl
CHAPTER SOUTION STOICHIOMETRY 111 Acid-Base Reactions 65. 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) 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) 66. 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) 67. 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.
11 CHAPTER SOUTION STOICHIOMETRY 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) 68. 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) c. Hydroiodic acid plus calcium hydroxide is a possibility. HI(aq) + Ca(OH) (aq) H O(l) + CaI (aq) 69. If we begin with 50.00 m of 0.00 M NaOH, then: 50.00 10 0.00 = 1.00 a. NaOH(aq) + HCl(aq) NaCl(aq) + H O(l) 1.00 10 NaOH 1HCl NaOH 10 NaOH is to be neutralized 1 0.100 b. HNO (aq) + NaOH(aq) H O(l) + NaNO (aq) 1.00 10 NaOH 1HNO NaOH c. HC H O (aq) + NaOH(aq) H O(l) + NaC H O (aq) 1.00 10 NaOH 1HCHO NaOH = 0.100 or 100. m 1 = 6.67 10 or 66.7 m 0.150 HNO 1 = 5.00 10 0.00 HCHO = 50.0 m 70. We begin with 5.00 m of 0.00 M HCl or 5.00 10 0.00 / = 5.00 10 HCl. a. HCl(aq) + NaOH(aq) H O(l) + NaCl(aq) 5.00 10 HCl 1HCl NaOH 1 = 5.00 0.100 NaOH 10 or 50.0 m b. HCl(aq) + Sr(OH) (aq) H O(l) + SrCl (aq)
CHAPTER SOUTION STOICHIOMETRY 11 5.00 10 HCl 1Sr(OH) HCl c. HCl(aq) + KOH(aq) H O(l) + KCl(aq) 1 = 5.00 10 0.0500Sr(OH) = 50.0 m 5.00 10 HCl 1KOH HCl 1 =.00 10 = 0.0 m 0.50 KOH 71. Ba(OH) (aq) + HCl(aq) BaCl (aq) + H O(l); H + (aq) + OH (aq) H O(l) 75.0 5.0 10 10 0.50 HCl = 1.88 10 HCl = 1.88 10 H + + 1.88 10 Cl 0.0550 Ba(OH) = 1. 10 Ba(OH) = 1. 10 Ba + +.8 10 OH The net ionic equation requires a 1 : 1 e ratio between OH and H +. The actual e OH to e H + ratio is greater than 1 : 1, so OH is in excess. Because 1.88 10 OH will be neutralized by the H +, we have (.8 1.88) 10 = 0.60 10 OH in excess. M = OH OH excess totalvolume 6.0 10 OH =.0 0.0750 0.50 10 M OH 7. 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 + = 0.0500 + 0.1000 Mol OH = 0.5000 + 0.000 0.100HCl 0.00 HNO 1H HCl 0.0100Ca(OH) 0.100 RbOH 1H = 0.00500 + 0.000 = 0.050 H + HNO OH Ca(OH) 1OH = 0.0100 + 0.000 = 0.000 OH RbOH We have an excess of OH, so the solution is basic (not neutral). The es of excess OH = 0.000 OH initially 0.050 OH reacted (with H + ) = 0.0050 OH excess. M = OH 0.0050OH (0.0500 0.1000 0.5000 0.000) 0.0050 0.8500 = 5.9 10 M
11 CHAPTER SOUTION STOICHIOMETRY 7. HCl(aq) + NaOH(aq) H O(l) + NaCl(aq).16 10 NaOH 0.106NaOH NaOH 1HCl =.56 10 HCl NaOH Molarity of HCl =.56 10 5.00 10 = 0.10 M HCl 7. HC H O (aq) + NaOH(aq) H O(l) + NaC H O (aq) a. 16.58 10 soln.506 NaOH soln 1HCHO NaOH 0 = 8.9 10 HC H O Concentration of HC H O (aq) = 8.910 0.01000 = 0.89 M b. If we have 1.000 of solution: Total mass = 1000. m 1.006g m = 1006 g solution Mass of HC H O = 0.89 60.05g = 50.0 g HC H O Mass % acetic acid = 50.0g 1006g 100 = 5.010% 75. HNO (aq) + Ca(OH) (aq) H O(l) + Ca(NO ) (aq) 5.00 10 HNO 0.0500 HNO HNO 1Ca(OH) HNO 1 Ca(OH) 0.000Ca(OH) = 0.08 =.8 m Ca(OH) 76. Strong bases contain the hydroxide ion (OH ). The reaction that occurs is H + + OH H O. 0.150 H 1OH 0.010 H = 1.80 10 OH The 0.0 m of the unknown strong base contains 1.80 10 OH. 1.80 10 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.
CHAPTER SOUTION STOICHIOMETRY 115 77. KHP is a monoprotic acid: NaOH(aq) + KHP(aq) H O(l) + NaKP(aq) Mass KHP = 0.006 NaOH 0.1000 NaOH NaOH 1KHP NaOH 0.g KHP KHP = 0.178 g KHP 78. 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 1KHP 0.g KHP 1 NaOH = 5.98 10 NaOH KHP There are 5.98 10 of sodium hydroxide in.67 m of solution. Therefore, the concentration of sodium hydroxide is: 5.98 10.67 10 = 1.58 10 M NaOH Oxidation-Reduction Reactions 79. 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. 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
116 CHAPTER SOUTION STOICHIOMETRY 80. 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 81. a. b. c. (x) + (+1) = 0, x = d. + e. +1 f. + g. + h. +5 i. 0 8. 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, + 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, + 8. 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.
CHAPTER SOUTION STOICHIOMETRY 117 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. 8. 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. 85. Each sodium atom goes from the 0 oxidation state in Na to the +1 oxidation state in NaF. Each Na atom loses one electron. Each fluorine atom goes from the 0 oxidation state in F to the 1 state in NaF. In order to match electrons gained by fluorine with electrons lost by sodium, 1 F atom is needed for every Na atom in the balanced equation. Because F contains two fluorine atoms, two sodium atoms will be needed to balance the electrons. The following balanced equation makes sense from an atom standpoint but also makes sense from an electron standpoint. Na(s) + F (g) NaF(s) 86. Each oxygen atom goes from the 0 oxidation state in O to the oxidation state in MgO. Each magnesium atom goes from the 0 oxidation state in Mg to the + oxidation state in MgO. To match electron gain with electron loss, 1 atom of O is needed for each atom of Mg in the balanced equation. Because two oxygen atoms are in each O ecule, we will need two Mg atoms for every O ecule. The balanced equation below balances atoms but also balances electrons, which must always be the case in any correctly balanced equation. Mg(s) + O (g) MgO(s) 87. a. The first step is to assign oxidation states to all atoms (see numbers above the atoms). +1 0 + +1 C H 6 + O CO + H O
118 CHAPTER SOUTION STOICHIOMETRY Each carbon atom changes from to +, an increase of 7. Each oxygen atom changes from 0 to, a decrease of. We need 7/ O atoms for every C atom in order to balance electron gain with electron loss. C H 6 + 7/ O CO + H O Balancing the remainder of the equation by inspection: or C H 6 (g) + 7/ O (g) CO (g) + H O(g) C H 6 (g) + 7 O (g) CO (g) + 6 H O(g) b. The oxidation state of magnesium changes from 0 to +, an increase of. The oxidation state of hydrogen changes from +1 to 0, a decrease of 1. We need H atoms for every Mg atom in order to balance the electrons transferred. The balanced equation is: Mg(s) + HCl(aq) Mg + (aq) + Cl (aq) + H (g) c. The oxidation state of nickel increases by (0 to +), and the oxidation state of cobalt decreases by 1 (+ to +). We need Co + ions for every Ni atom in order to balance electron gain with electron loss. The balanced equation is: Ni(s) + Co + (aq) Ni + (aq) + Co + (aq) d. The equation is balanced (mass and charge balanced). Each hydrogen atom gains one electron (+1 0), and each zinc atom loses two electrons (0 +). We need H atoms for every Zn atom in order to balance the electrons transferred. This is the ratio in the given equation: Zn(s) + H SO (aq) ZnSO (aq) + H (g) 88. a. The first step is to assign oxidation states to all atoms (see numbers above the atoms). 0 0 + 1 Cl + Al Al + + Cl Each aluminum atom changes in oxidation state from 0 to +, an increase of. Each chlorine atom changes from 0 to 1, a decrease of 1. We need Cl atoms for every Al atom in the balanced equation in order to balance electron gain with electron loss. / Cl + Al Al + + Cl For whole numbers, multiply through by two. The balanced equation is: Cl (g) + Al(s) Al + (aq) + 6 Cl (aq) 0 +1 0 + +1 b. O + H O + Pb Pb(OH) From the oxidation states written above the elements, lead is oxidized, and oxygen in O is reduced. Each lead atom changes from 0 to +, an increase of, and each O atom in
CHAPTER SOUTION STOICHIOMETRY 119 O changes from 0 to, a decrease of. We need 1 Pb atom for each O atom in O to balance the electrons transferred. Balancing the electrons: O + H O + Pb Pb(OH) The last step is to balance the rest of the equation by inspection. In this reaction, when the H atoms become balanced, the entire equation is balanced. The balanced overall equation is: O (g) + H O(l) + Pb(s) Pb(OH) (s) +1 +7 + + + +1 c. H + + MnO + Fe + Mn + + Fe + + H O From the oxidation states written above each element, manganese is reduced (goes from +7 to +), and Fe is oxidized (goes from + to +). In order to balance the electrons transferred, we need 5 Fe atoms for every Mn atom. Balancing the electrons gives: H + + MnO + 5 Fe + Mn + + 5 Fe + + H O Balancing the O atoms, then the H atoms by inspection, leads to the following overall balanced equation. 8 H + (aq) + MnO (aq) + 5 Fe + (aq) Mn + (aq) + 5 Fe + (aq) + H O(l) Additional Exercises 89. 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 and 0.00 of KIO. Molar mass of KIO = 9.10 + 16.9 + (16.00) = 1.0 g/ 0.0198 1.0 g =.7 g; 0.00 1.0 g =. 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. 90. Solution A: ecules 1.0 ; solution B: 6 ecules.0 1.5 ecules 1.0 Solution C: ecules ecules ; solution D:.0 1.0 6 ecules.0 ecules 1.0
10 CHAPTER SOUTION STOICHIOMETRY Solution A has the most ecules per unit volume so solution A is most concentrated. This is followed by solution D, then solution C. Solution B has the fewest ecules per unit volume, so solution B is least concentrated. 91..0 g C 1 H O 11 1 11 C1 HO = 0.095 C 1 H O 11 added to blood.0g The blood sugar level would increase by: 0 11.095C1 HO = 0.019 / 5.0 9. Mol CaCl present = 0.0 CaCl 0.75CaCl CaCl The volume of CaCl solution after evaporation is: 1 CaCl 6. 10 CaCl 1.10CaCl = 6. 10 CaCl = 5.75 10 = 57.5 m CaCl Volume H O evaporated = 0. m 57.5 m = 17 m H O evaporated 9. 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)
CHAPTER SOUTION STOICHIOMETRY 11 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) 9. a. MgCl (aq) + AgNO (aq) AgCl(s) + Mg(NO ) (aq) 0.61 g AgCl 1AgCl 1. g AgCl 0.1g MgCl 100 = 1.% MgCl 1.50g mixture 1MgCl 95.1g = 0.1 g MgCl AgCl MgCl b. 0.1 g MgCl 1MgCl 95.1g AgNO MgCl 1 0.500 AgNO 1000m 1 95. XCl (aq) + AgNO (aq) AgCl(s) + X(NO ) (aq) = 8.95 m AgNO 1.8 g AgCl AgCl 1. g 1XCl =.81 AgCl 1 10 XCl.91 10 1.00g XCl XCl = 08 g/; x + (5.5) = 08, x = 17 g/ From the periodic table, the metal X is barium (Ba). 96. From the periodic table, use aluminum in the formulas to convert from mass of Al(OH) to mass of Al (SO ) in the mixture. 0.107 g Al(OH) 1Al(OH) 78.00g 1Al Al(OH) 1Al(SO Al ).17g Al(SO ) = 0.5 g Al (SO ) Al (SO ) Mass % Al (SO ) = 0.5g 1.5g 100 = 16.% 97. All the Tl in TlI came from Tl in Tl SO. The conversion from TlI to Tl SO uses the ar 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 = 0.190g Tl SO 100 = 1.65% Tl SO 9.86g pesticide
1 CHAPTER SOUTION STOICHIOMETRY 98. a. Fe + (aq) + OH (aq) Fe(OH) (s) Fe(OH) : 55.85 + (16.00) + (1.008) = 106.87 g/ 0.107 g Fe(OH) 55.85g Fe 106.87g Fe(OH) = 0.0559 g Fe b. Fe(NO ) : 55.85 + (1.01) + 9(16.00) = 1.86 g/ 0.0559 g Fe 1.86g Fe(NO ) 55.85g Fe = 0. g Fe(NO ) c. Mass % Fe(NO ) = 0.g 0.56g 100 = 5.1% 99. 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 1Cr(OH) 10.0g NaOH = 6.00 10 Cr(OH) NaOH(aq) + HCl(aq) NaCl(aq) + H O(l) Mol NaOH used = 0.1000 to react with HCl 0.00 HCl 1 NaOH =.00 10 HCl M NaOH = total NaOH volume = 6.00 10.00 10 0.0500 =.00 M NaOH 100. a. MgO(s) + HCl(aq) MgCl (aq) + H O(l) Mg(OH) (s) + HCl(aq) MgCl (aq) + H O(l) Al(OH) (s) + HCl(aq) AlCl (aq) + H O(l) b. et's calculate the number of es of HCl neutralized per gram of substance. We can get these directly from the balanced equations and the ar masses of the substances. HCl MgO 1MgO 0.1g MgO.9610 HCl g MgO HCl Mg(OH) 1Mg(OH).910 58.g Mg(OH) g Mg(OH) HCl
CHAPTER SOUTION STOICHIOMETRY 1 HCl Al(OH) 1Al(OH).8610 78.00g Al(OH) g Al(OH) HCl Therefore, 1 gram of magnesium oxide would neutralize the most 0.10 M HCl. 101. Using HA as an abbreviation for the monoprotic acid acetylsalicylic acid: HA(aq) + NaOH(aq) H O(l) + NaA(aq) 0.5065 NaOH 1HA Mol HA = 0.0517 NaOH = 1.781 10 HA NaOH NaOH Fom the problem,.10 g HA was reacted, so:.10g HA ar mass = = 180. g/ 1.781 10 HA 10. Mg(s) + HCl(aq) MgCl (aq) + H (g).00 g Mg 1Mg.1g Mg HCl 1 = 0.09 = 9 m HCl Mg 5.0 HCl 10. et HA = unknown monoprotic acid; HA(aq) + NaOH(aq) NaA(aq) + H O(l) Mol HA present = 0.050 0.500 NaOH 1HA 1 NaOH = 0.015 HA x g HA HA.0g HA 0.015 HA, x = ar mass of HA = 176 g/ Empirical formula weight (1) + (1) + (16) = 88 g/. Because 176/88 =.0, the ecular formula is (C H O ) = C 6 H 8 O 6. 10. We get the empirical formula from the elemental analysis. Out of 100.00 g carminic acid, there are: 5.66 g C 1 C 1.01g C =.68 C;.09 g H 1 H 1.008g H =.06 H.5 g O 1 O 16.00g O =.61 O Dividing the es by the smallest number gives:.68 1.69;.61.06.61 1.5
1 CHAPTER SOUTION STOICHIOMETRY These numbers don t give obvious e ratios. et s determine the C to H ratio:.68.06. 06 So let's try 10 1.10 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 0 O 1 is the empirical formula. We can get ar 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 + ). 18.0 10 soln Molar mass = 7. 10 0.006 NaOH 1 carminicacid soln NaOH 0.60g 9g = 7. 10 carminic acid The empirical formula mass of C H 0 O 1 (1) + 0(1) + 1(16) = 9 g. Therefore, the ecular formula of carminic acid is also C H 0 O 1. 105. 0.10 g AgCl 1AgCl 1.g AgCl 1Cl AgCl 5.5g Cl Cl =.57 10 g 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 1 H 0 Cl 6 N which contains.57 10 g Cl. Molar mass of C 1 H 0 Cl 6 N = 1(1.01) + 0(1.008) + 6(5.5) + (1.01) = 9.0 g/ There are 6(5.5) = 1.70 g chlorine for every e (9.0 g) of C 1 H 0 Cl 6 N. 9.0g C.57 10 g Cl 1H0Cl 6N 1.70g Cl Mass % chlorisondamine chloride = 5.18 10 1.8g = 5.18 10 g C 1 H 0 Cl 6 N g 100 =.05% 106. All the sulfur in BaSO came from the saccharin. The conversion from BaSO to saccharin utilizes the ar masses of each compound. 0.50 g BaSO.07g S 18.19g C7H5NOS = 0.99 g C 7 H 5 NO S. g BaSO.07g S Averagemass Tablet 0.99g 10 tablets.99 10 tablet g 9.9mg tablet
CHAPTER SOUTION STOICHIOMETRY 15 Average mass % = 0.99g C7H5NO S 0.589g 100 = 67.00% saccharin by mass 107. Use the silver nitrate data to calculate the Cl present, then use the formula of douglasite (KCl FeCl H O) to convert from Cl to douglasite (1 e of douglasite contains es of Cl ). The net ionic equation is Ag + + Cl AgCl(s). 0.070 Mass % douglasite = 0.1000 Ag 1Cl Ag 0.900g 100 = 6.7% 0.550g 1 douglasite Cl 11.88g douglasite = 0.900 g douglasite 108. a. Al(s) + HCl(aq) AlCl (aq) + / H (g) or Al(s) + 6 HCl(aq) AlCl (aq) + Hydrogen is reduced (goes from the +1 oxidation state to the 0 oxidation state), and aluminum Al is oxidized (0 +). H (g) b. Balancing S is most complicated since sulfur is in both products. Balance C and H first; then worry about S. CH (g) + S(s) CS (l) + H S(g) Sulfur is reduced (0 ), and carbon is oxidized ( +). c. Balance C and H first; then balance O. C H 8 (g) + 5 O (g) CO (g) + H O(l) Oxygen is reduced (0 ), and carbon is oxidized (8/ +). d. Although this reaction is mass balanced, it is not charge balanced. We need es of silver on each side to balance the charge. Cu(s) + Ag + (aq) Ag(s) + Cu + (aq) Silver is reduced (+1 0), and copper is oxidized (0 +). 109. Cr O 7 : (x) + 7() =, x = +6 C H 5 OH (C H 6 O): (y) + 6(+1) + () = 0, y = CO : z + () = 0, z = + Each chromium atom goes from the oxidation state of +6 in Cr O 7 to + in Cr +. Each chromium atom gains three electrons; chromium is the species reduced. Each carbon atom goes from the oxidation state of in C H 5 OH to + in CO. Each carbon atom loses six
16 CHAPTER SOUTION STOICHIOMETRY electrons; carbon is the species oxidized. From the balanced equation, we have four chromium atoms and two carbon atoms. With each chromium atom gaining three electrons, a total of () = 1 electrons are transferred in the balanced reaction. This is confirmed from the carbon atoms in the balanced equation, where each carbon atom loses six electrons [(6) = 1 electrons transferred]. ChemWork Problems The answers to the problems 110-119 (or a variation to these problems) are found in OW. These problems are also assignable in OW. Challenge Problems 10. et x = mass of NaCl, and let y = mass K SO. So x + y = 10.00. Two reactions occur: Pb + (aq) + Cl (aq) PbCl (s) and Pb + (aq) + SO (aq) PbSO (s) Molar mass of NaCl = 58. g/; ar mass of K SO = 17.7 g/; ar mass of PbCl = 78.1 g/; ar mass of PbSO = 0. g/ x 58. = es NaCl; y 17.7 = es K SO mass of PbCl + mass PbSO = total mass of solid x y (1/)(78.1) + (0.) = 1.75 58. 17.7 We have two equations: (.79)x + (1.70)y = 1.75 and x + y = 10.00. Solving: x = 6.81 g NaCl; 6.81g NaCl 10.00g mixture 100 = 68.1% NaCl 11. a. 5.0 ppb Hg in water 5.0 ng Hg g soln 9 5.0 10 g Hg msoln 9 5.0 10 g Hg m 1Hg 00.6g Hg 1000m =.5 8 10 M Hg b. 9 1.0 10 g CHCl m 1CHCl 119.7g CHCl 1000m = 8. 9 10 M CHCl 10.0 μg As 10.0 10 g As c. 10.0 ppm As g soln msoln 6