Solution Stoichiometry

Transcription

1 Chapter 8 Solution Stoichiometry Note to teacher: You will notice that there are two different formats for the Sample Problems in the student textbook. Where appropriate, the Sample Problem contains the full set of steps: Problem, What Is Required, What Is Given,, Act on Your Strategy, and. Where a shorter solution is appropriate, the Sample Problem contains only two steps: Problem and Solution. Where relevant, a Check Your Solution step is also included in the shorter Sample Problems. Solutions for Practice Problems Student Textbook page Problem Decide whether each of the following salts is soluble or insoluble in distilled water. Give reasons for your answer. (a) lead(ii) chloride, PbCl 2 (a white crystalline powder used in paints) (b) zinc oxide, ZnO (a white pigment used in paints, cosmetics, and calamine lotion) (c) silver acetate, AgCH 3 COO (a whitish powder that is used to help people quit smoking because of the bitter taste it produces) What Is Required? Determine if the salts listed will dissolve in water. What Is Given? You have the names of the three salts and access to the solubility guidelines. Identify the cations and anions in each salt. Refer to the solubility chart in the student textbook and determine the guideline solubility of these ions. The solubility of the salt will correspond with the solubility of the ion having the higher guideline (lower number). (a) Salts of chloride are soluble, and lead (II) chloride is not an exception. (b) The solubility of oxides is not given in the chart, but students may recognize that zinc oxide is insoluble. (c) Most salts of acetates are soluble, but silver acetate is an exception. Check your results against another reference. These conclusions are correct. 2. Problem Which of the following compounds are soluble in water? Explain your reasoning for each compound. (a) potassium nitrate, KNO 3 (used to manufacture gunpowder) (b) lithium carbonate, Li 2 CO 3 (used to treat people who suffer from depression) (c) lead(ii) oxide, PbO (used to make crystal glass) What Is Required? Determine if the salts listed will dissolve in water. 145

2 What Is Given? You have the names of the three salts and access to the solubility guidelines. Identify the cations and anions in each salt. Refer to the solubility chart in the student textbook and determine the guideline solubility of these ions. The solubility of the salt will correspond with the solubility of the ion having the higher guideline (lower number). (a) All salts of nitrate and potassium are soluble. Therefore, potassium nitrate is soluble. (b) All salts of Group 1 cations, including lithium, are soluble. Therefore, lithium carbonate is soluble. (c) The solubility of oxides is not included in the solubility table. Students will likely predict that lead(ii) oxide is insoluble since other compounts of lead seem to be insoluble. Check the predictions against another reference. These results are correct. 3. Problem Which of the following compounds are insoluble in water? (a) calcium carbonate (present in marble and limestone) (b) magnesium sulfate, MgSO 4 (found in the hydrated salt, MgSO 4 7H 2 O, also known as Epsom salts; used for the relief of aching muscles and as a laxative) (c) aluminum phosphate, AlPO 4 (found in dental cements) What Is Required? Determine if the salts listed will dissolve in water. What Is Given? You have the names of the three salts and access to the solubility guidelines. Identify the cations and anions in each salt. Refer to the solubility chart in the student textbook and determine the guideline solubility of these ions. The solubility of the salt will correspond with the solubility of the ion having the higher guideline (lower number). Name Formula Ions Guideline Ion Solubility Salt Solubility calcium carbonate CaCO 3 Ca 2+ 2 CO insoluble insoluble INSOLUBLE magnesium sulfate MgSO 4 Mg 2+ 2 SO soluble soluble SOLUBLE aluminum phosphate AlPO 4 Al 3+ 3 PO soluble insoluble INSOLUBLE Check the predictions against another reference. These results are correct. 146

3 Solutions for Practice Problems Student Textbook page Problem Predict the result of mixing each pair of aqueous solutions. Write a balanced chemical equation if you predict that a precipitate forms. Write NR if you predict that no reaction takes place. (a) sodium sulfide and iron(ii) sulfate (b) sodium hydroxide and barium nitrate (c) cesium phosphate and calcium bromide (d) sodium carbonate and sulfuric acid (e) sodium nitrate and copper(ii) sulfate (f) ammonium iodide and silver nitrate (g) potassium carbonate and iron(ii) nitrate (h) aluminum nitrate and sodium phosphate (i) potassium chloride and iron(ii) nitrate (j) ammonium sulfate and barium chloride (k) sodium sulfide and nickel(ii) sulfate (l) lead(ii) nitrate and potassium bromide What Is Required? Predict whether or not each pair of aqueous solutions will provide ions that will combine to form an insoluble product (precipitate). Write a balanced chemical equation if a reaction is predicted and NR if no reaction is predicted. What Is Given? You know the names of the compound in each solution. Identify the ions in each compound. Exchange the cations in the two compounds and for each new compound, look up its solubility guideline in the student textbook. Predict whether or not either of these new compounds is insoluble. Write the balanced equation for those examples in which a new insoluble product formed and NR if both the new compounds are soluble. Solubility Key: S = soluble I = insoluble (a) sodium sulfide + iron(ii) nitrate Reactant Ions Na + S 2- Fe SO 4 Products Ions Na + 2- SO 4 Fe 2+ S 2- Guideline Ion Solubility S S S I Product Na 2 SO 4 FeS Product Solubility S I Na 2 S (aq) + FeSO 4(aq) Na 2 SO 4(aq) + FeS (s) (b) sodium hydroxide + barium nitrate Reactant Ions Na + OH - Ba 2+ - NO 3 Products Ions Na + - NO 3 Ba 2+ OH - Guideline Ion Solubility S S I I Product NaNO 3 Ba(OH) 2 147

4 Product Solubility S S (exception to rule) NaOH (aq) + Ba(NO 3 ) 2(aq) NR (c) cesium phosphate + calcium bromide Reactant Ions Cs + 3- PO 4 Ca 2+ Br - Products Ions Cs + Br - Ca PO 4 Guideline Ion Solubility S S I I Product CsBr Ca 3 PO 4 Product Solubility S I 2Cs 3 PO 4(aq) + 3CaBr 2(aq) 6CsBr (aq) + Ca 3 (PO 4 ) 2(s) (d) sodium carbonate and sulfuric acid Reactant Ions Na + 2- CO 3 H + 2- SO 4 Products Ions Na + 2- SO 4 H + 2- CO 3 Guideline Ion Solubility S S S I Product Na 2 SO 4 H 2 CO 3 Product Solubility S S Na 2 CO 3(aq) + H 2 SO 4(aq) NR (e) sodium nitrate and copper(ii) sulfate Reactant Ions Na + - NO 3 Cu SO 4 Products Ions Na + 2- SO 4 Cu 2+ - NO 3 Guideline Ion Solubility S S S S Product Na 2 SO 4 Cu(NO 3 ) 2 Product Solubility S S NaNO 3(aq) + CuSO 4(aq) NR (f) ammonium iodide and silver nitrate Reactant Ions + NH 4 I - Ag + - NO 3 Products Ions + NH 4 - NO 3 Ag + I - Guideline Ion Solubility S S I S Product NH 4 NO 3 AgI Product Solubility S S NH 4 I (aq) + AgNO 3(aq) NH 4 NO 3(aq) + AgI (s) (g) potassium carbonate and iron(ii) nitrate Reactant Ions K + 2- CO 3 Fe 2+ - NO 3 Products Ions K + - NO 3 Fe CO 3 Guideline Ion Solubility S S S I Product KNO 3 FeCO 3 Product Solubility S I K 2 CO 3(aq) + Fe(NO 3 ) 2(aq) 2KNO 3(aq) + FeCO 3(s) (h) aluminum nitrate and sodium phosphate Reactant Ions Al 3+ - NO 3 Na + 3- PO 4 Products Ions Al PO 4 Na + - NO 3 148

5 Guideline Ion Solubility S I S S Product AlPO 4 NaNO 3 Product Solubility I I Al(NO 3 ) 3(aq) + Na 3 PO 4(aq) 3NaNO 3(aq) + AlPO 4(s) (i) potassium chloride and iron(ii) Reactant Ions K + Cl - Fe 2+ - NO 3 Products Ions K + - NO 3 Fe 2+ Cl - Guideline Ion Solubility S S S S Product KNO 3 FeCl 2 Product Solubility S S KCl (aq) + Fe(NO 3 ) 2(aq) NR (j) ammonium sulfate and barium chloride Reactant Ions + NH 4 2- SO 4 Ba 2+ Cl - Products Ions + NH 4 Cl - Ba SO 4 Guideline Ion Solubility S S I S Product NH 4 Cl BaSO 4 Product Solubility S I (NH 4 ) 2 SO 4(aq) + BaCl 2(aq) 2NH 4 Cl (aq) + BaSO 4(s) (k) sodium sulfide and nickel(ii) sulfate Reactant Ions Na + S 2- Ni SO 4 Products Ions Na + 2- SO 4 Ni 2+ S 2- Guideline Ion Solubility S S S I Product Na 2 SO 4 NiS Product Solubility S I Na 2 S (aq) + NiSO 4(aq) Na 2 SO 4(aq) + NiS (s) (l) lead(ii) nitrate and potassium bromide Reactant Ions Pb 2+ - NO 3 K + Br - Products Ions Pb 2+ Br - K + - NO 3 Guideline Ion Solubility I S S S Product PbBr 2 KNO 3 Product Solubility I S Pb(NO 3 ) 2(aq) + 2KBr (aq) 2KNO 3(aq) + PbBr 2(s) Examine the final equation to see if the ionic compounds that are noted as being in aqueous solution are soluble and the compounds noted as solid are insoluble. 149

6 Solutions for Practice Problems Student Textbook page Problem Mixing each pair of aqueous solutions results in a chemical reaction. Identify the spectator ions. Then write the balanced net ionic equation. (a) sodium carbonate and hydrochloric acid (b) sulfuric acid and sodium hydroxide What Is Required? Identify the spectator ions and write a balanced net ionic equation. What Is Given? You know the chemical names of the compounds. Write the chemical formula of each compound, then complete the chemical equation for the reaction. Using the solubility guideline, note which product is soluble and which is the precipitate. Replace the formulae of the soluble ionic compounds with dissociated ions. Identify the ions that appear on both sides of the equation as spectator ions. Rewrite the equation without the spectator ions. (a) Na 2 CO 3(aq) + 2HCl (aq) 2NaCl (aq) + H 2 CO 3(aq) H 2 CO 3(aq) CO 2(g) + H 2 O (l) 2Na + 2- (aq) + CO 3 (aq) + 2H + (aq) + 2Cl - (aq) 2Na + (aq) + 2Cl - (aq) + CO 2(g) + H 2 O (l) spectator ions are: Na + (aq) and Cl - (aq) net ionic equation: 2H + 2- (aq) + CO 3 (aq) CO 2(g) + H 2 O (l) (b) H 2 SO 4(aq) + 2NaOH (aq) Na 2 SO 4(aq) + 2H 2 O (l) 2H + 2- (aq) + SO 4 (aq) + 2Na + (aq) + 2OH - (aq) 2Na + 2- (aq) + SO 4 (aq) + 2H 2 O (l) spectator ions are: Na + 2- (aq) and SO 4 (aq) net ionic equation: 2H + (aq) + OH - (aq) H 2 O (l) Examine the net ionic equation to confirm that no ions are common to both sides and that the equation is balanced. 6. Problem Identify the spectator ions for the reaction that takes place when each pair of aqueous solutions is mixed. Then write the balanced net ionic equation. (a) ammonium phosphate and zinc sulfate (b) lithium carbonate and nitric acid (c) sulfuric acid and barium hydroxide What Is Required? Identify the spectator ions and write a balanced net ionic equation. What Is Given? You know the chemical names of the compounds. 150

7 Write the chemical formula of each compound, then complete the chemical equation for the reaction. Using the solubility guideline, note which product is soluble and which is the precipitate. Replace the formulae of the soluble ionic compounds with dissociated ions. Identify the ions that appear on both sides of the equation as spectator ions. Rewrite the equation without the spectator ions. (a) 2(NH 4 ) 3 PO 4(aq) + 3ZnSO 4(aq) Zn 3 (PO 4 ) 2(s) + 3(NH 4 ) 2 SO 4(aq) + 3-6NH 4 (aq) + 2PO 4 (aq) + 3Zn (aq) + 3SO 4 (aq) 3Zn (aq) + 2PO 4 (aq) NH 4 (aq) + 3SO 4 (aq) + 2- spectator ions are: NH 4 (aq) and SO 4 (aq) net ionic equation: 3Zn (aq) + 2PO 4 (aq) Zn 3 (PO 4 ) 2(s) (b) Li 2 CO 3(aq) + 2HNO 3(aq) 2LiNO 3(aq) + H 2 O (l) + CO 2(g) 2Li + 2- (aq) + CO 3 (aq) + 2H + - (aq) + 2NO 3 (aq) 2Li + - (aq) + 2NO 3 (aq) + H 2 O (l) +CO 2(g) spectator ions are: Li + - (aq) and NO 3 (aq) net ionic equation: 2H + 2- (aq) + CO 3 (aq) CO 2(g) + H 2 O (l) (c) H 2 SO 4(aq) + Ba(OH) 2(aq) BaSO 4(s) + 2H 2 O (l) 2H + 2- (aq) + SO 4 (aq) + Ba 2+ (aq) + 2OH - (aq) BaSO 4(s) + 2H 2 O (l) spectator ions are: none net ionic equation: 2H + 2- (aq) + SO 4 (aq) + Ba 2+ (aq) + 2OH - (aq) BaSO 4(s) + 2H 2 O (l) Examine the net ionic equation to confirm that no ions are common to both sides and that the equation is balanced. Solutions for Practice Problems Student Textbook page Problem Write dissociation equations for the following ionic compounds in water, and state the concentration of the dissociated ions in each aqueous solution. Assume that each compound dissolves completely. (a) 1 mol/l lithium fluoride (b) 0.5 mol/l sodium bromide (c) 1.5 mol/l sodium sulfide (d) 1 mol/l calcium nitrate You must indicate the ions that form when the ionic compound dissociates and the concentration of each ion in mol/l. You know the name of the ionic compound and its concentration. Write the chemical formula for each ionic compound. Recall that the chemical formula indicates the kind of number of ions in the compound. Refer to the student textbook and periodic table for the correct formula and charge on the ions. The equation must be balance atomically and by charge. This means that the total charge on the positive ion(s) must balance the total charge on the negative ions. The mol ratio in the balanced dissociation equation will determine the concentration of each ion. 151

8 (a) + - LiF(s) Æ Li ( aq) + F (aq) 1 mol / L 1 mol / L 1 mol / L (b) (c) + - NaBr(s) Æ Na ( aq) + Br (aq) 05. mol / L 0.5 mol / L 0.5 mol / L + 2- Na2 S(s) Æ 2Na ( aq) + S (aq) 15. mol / L 3.0 mol / L 1.5 mol / L (d) 2+ - Ca(NO3 ) 2(s) Æ Ca ( aq) + 2NO3 (aq) 10. mol / L 1.0 mol / L 2. 0mol / L In each case the total charge on the positive ions numerically balances the total on the negative ions. The concentration of the ions follows the mol ratio in the balanced dissociation equation. 8. Problem Write dissociation equations for the following compounds, and state the concentration of the dissociated ions in each aqueous solution. Assume that each compound dissolves completely. (a) 1.75 mol/l ammonium chloride (b) 0.25 mol/l ammonium sulfide (c) 6 mol/l ammonium sulfate (d) 2 mol/l silver nitrate You must indicate the ions that form when the ionic compound dissociates and the concentration of each ion in mol/l. You know the name of the ionic compound and its concentration. Write the chemical formula for each ionic compound. Recall that the chemical formula indicates the kind of number of ions in the compound. Refer to the student textbook and periodic table for the correct formula and charge on the ions. The equation must be balance atomically and by charge. This means that the total charge on the positive ion(s) must balance the total charge on the negative ions. The mol ratio in the balanced dissociation equation will determine the concentration of each ion. + - (a) NH4 Cl(s) Æ NH4 ( aq) + Cl (aq) 175. mol/l 1.75 mol/l 175. mol/l (b) (NH ) S Æ 2NH + S (s) 4 ( aq) (aq) 025. mol/l 0.50 mol/l 025. mol/l (c) (NH ) SO Æ 2NH + SO (s) 4 ( aq) 4 (aq) 60. mol/l 12.0 mol/l 60. mol/l + - (d) AgNO3(s) Æ Ag ( aq) + NO3 (aq) 20. mol/l 2.0 mol/l 20. mol/l 152

9 In each case the total charge on the positive ions numerically balances the total on the negative ions. The concentration of the ions follows the mol ratio in the balanced dissociation equation. 9. Problem 18.2 g of CaCl 2 is dissolved in ml of water to make a clear, colourless solution. Give the dissociation equation for this physical change, and state the concentration of the dissociated ions. You must indicate the ions that form when the ionic compound dissociates and the concentration of each ion in mol/l. You know the name of the ionic compound and its concentration. Write the chemical formula for calcium chloride and determine its molar mass. Use m the formula n = to find the moles of calcium chloride. Use the formula C = n M V to determine the molar concentration. Write the dissociation equation and use the mol ratio to indicate the concentration of each ion. M for CaCl 2 = 111 g/mol 1 mol 18.2 g CaCl 2 = mol CaCl g n mol C = = =0.547 mol / L V L CaCl2(s) Æ Ca ( aq) + 2Cl (aq) mol/l mol/l mol/l The equation is balanced atomically and by charge and the concentration of the ions follows the mol ratio in this balanced dissociation equation. 10. Problem Cobalt(II) chloride is a pale blue compound that dissolves easily in water. It is often used to detect excess humidity in air, because the compound dissolves and turns pink. (a) 14.2 g of cobalt(ii) chloride is dissolved in 1.50 L of water. Determine the concentration of CoCl 2(aq). (b) Write the dissociation equation for this change. (c) Determine the concentration of all ions in the final solution. You must indicate the ions that form when the ionic compound dissociates and the concentration of each ion in mol/l. You know the name of the ionic compound and its concentration. Write the chemical formula for cobalt(ii) chloride and determine its molar mass. Use m the formula n = to find the moles of calcium chloride. Use the formula C = n M V to determine the molar concentration. Write the dissociation equation and use the mol ratio to indicate the concentration of each ion. 153

10 (a) M for CoCl 2 = g/mol 14.2 g CoCl 2 1 mol =.109 mol CoCl g n mol C = = = mol / L V L (b) and (c) CoCl2(s) Æ Co ( aq) + 2Cl (aq) mol/l mol/l mol/ L The equation is balanced atomically and by charge and the concentration of the ions follows the mol ratio in this balanced dissociation equation. Solutions for Practice Problems Student Textbook page Problem Equal volumes of mol/l lithium nitrate and mol/l sodium nitrate are mixed together. Determine the concentration of nitrate ions in the mixture. You must calculate the concentration of the NO 3 (aq) in a mixture of two ionic compounds. The name and concentration of the two ionic compounds are given. Write the chemical formula for each compound, write the dissociation equation and use the mole ratio in this balanced equation to determine the concentration of the NO 3 (aq). Let the volume of each solution = V. The number of moles of NO 3 (aq) can be calculated from n = C V. The final concentration of the NO 3 (aq) can be calculated as + - LiNO Æ Li NO 3(s) ( aq) + 3 (aq) mol / L mol / L mol / L + - NaNO3(s) Æ Na ( aq) + NO mol / L mol / L mol / L total moles of ion final concentration= total volume _ mol / L V mol/ L V final [NO3 (aq)] = = mol / L 2 V When mixing equal volumes of two aqueous solutions it is reasonable that the final concentration will be an average of that of the two original concentrations. This answer supports that reasoning. 154

11 12. Problem Equal volumes of 1.25 mol/l magnesium sulfate and mol/l ammonium sulfate are mixed together. What is the concentration of sulfate ions in the mixture? You must calculate the concentration of the SO 4 2 (aq) in a mixture of two ionic compounds. The name and concentration of the two ionic compounds are given. Write the chemical formula for each compound, write the dissociation equation and use the mole ratio in this balanced equation to determine the concentration of the SO 2 4 (aq). Let the volume of each solution = V. The number of moles of SO 2 4 (aq) can be calculated from n = C V. The final concentration of the SO 2 4 (aq) can be calculated as total moles of ion. total volume MgSO4(s) Æ Mg ( aq) + SO4 (aq) 125. mol / L 1.25 mol / L 125. mol / L (NH 4) 2SO4(s) Æ 2NH4 ( aq) + SO4 (aq) mol / L mol / L mol / L total moles of ion final concentration = total volume mol / L V mol/ L V final [SO4 (aq)] = 2 V mol/ L When mixing equal volumes of two aqueous solutions it is reasonable that the final concentration will be an average of that of the two original concentrations. This answer supports that reasoning. 13. Problem A solution is prepared by adding 25.0 ml of mol/l sodium chloride with 35.0 ml of mol/l barium chloride. What is the concentration of chloride ions in the resulting solution, assuming that the volumes are additive? You must calculate the concentration of the Cl (aq) in a mixture of two ionic compounds. The name and concentration of the two ionic compounds and the volume of each solution are given. Write the chemical formula for each compound, write the dissociation equation and use the mole ratio in this balanced equation to determine the concentration of the Cl (aq). The number of moles of Cl (aq) can be calculated from n = C V. The final concentration of the Cl (aq) can be calculated as total moles of ion total volume = 155

12 + - NaCl(s) Æ Na ( aq) + Cl (aq) mol / L mol / L mol / L BaCl2(s) Æ Ba ( aq) + 2Cl (aq) mol / L mol / L mol / L total moles of ion final concentration = total volume mol / L L mol / L L final [Cl ] = = mol/ L L The final concentration of the Cl (aq) is expected to be a weighted average of the two original concentrations. A larger volume and a higher concentration of Cl (aq) comes from the BaCl 2 solution and therefore the final concentration of Cl (aq) should be closer to the original concentration of this solution. This is the case for the answer obtained. 14. Problem What is the concentration of nitrate ions in a solution that results by mixing 350 ml of mol/l potassium nitrate with 475 ml of mol/l magnesium nitrate? You must calculate the concentration of the NO 3 (aq) in a mixture of two ionic compounds. The name and concentration of the two ionic compounds and the volume of each solution are given. Write the chemical formula for each compound, write the dissociation equation and use the mole ratio in this balanced equation to determine the concentration of the NO 3 (aq). The number of moles of NO 3 (aq) can be calculated from n = C V. The final concentration of the NO 3 (aq) can be calculated as total moles of ion. total volume + KNO Æ K NO 3(s) ( aq) + 3 (aq) mol / L mol / L mol / L 2 + Mg(NO3 ) 2(s) Æ Mg ( aq) + 2NO3 (aq) mol / L mol / L mol / L total moles of ion final concentration = total volume mol / L L mol/ L L final [NO3 (aq)] = = mol L The final concentration of the NO 3 (aq) is expected to be a weighted average of the two original concentrations. A larger volume and a higher concentration of NO 3 (aq) comes from the Mg(NO 3 ) 2 solution and therefore the final concentration of NO 3 (aq) should be closer to the original concentration of this solution. This is the case for the answer obtained. 156

13 Solutions for Practice Problems Student Textbook page Problem Food manufacturers sometimes add calcium acetate to puddings and sweet sauces as a thickening agent. What volume of mol/l calcium acetate, Ca(CH 3 COO) 2(aq), contains mol of acetate ions? What Is Required? Find the volume of mol/l calcium acetate, Ca(CH 3 COO) 2(aq), solution that contains mol of acetate ions? What Is Given? You know the concentration of the calcium acetate solution, the formula for calcium acetate, and the required number of moles of acetate ion. Write the equation for the dissociation of Ca(CH 3 COO) 2. Determine the ratio of moles CH 3 COO - (aq) : Ca(CH 3COO) 2 to calculate the moles of Ca(CH 3 COO) 2(aq). Rearrange the formula n = C V and calculate the volume of solution required. Ca(CH 3 COO) 2 Ca 2+ (aq) + 2CH 3 COO - (aq) CH 3 COO - 1molCa(CH 3COO) 2 = mol Ca(CH 2molCH 3 COO - 3 COO) 2 V = = mol = L or 300 ml of Ca(CH COO) n C The final answer has the correct unit and number of significant figures. The answer seems to be reasonable. 16. Problem In a neutralization reaction, hydrochloric acid reacts with sodium hydroxide to produce sodium chloride and water: HCl (aq) + NaOH (aq) Æ NaCl (aq) + H 2 O What is the minimum volume of mol/l HCl that is needed to combine exactly with 22.7 ml of mol/l aqueous sodium hydroxide? You must find the volume of HCl that reacts with the NaOH. You know a volume and concentration of NaOH and a concentration of HCl. Calculate the mol of NaOH using n = C V. Use the mol ratio in the balanced equation to calculate the mol of HCl that combines with this number of mol of NaOH. Use the formula mol/l 3 2(aq) V = n C to calculate the volume of HCl. mol NaOH = mol/l L = mol NaOH and HCl react 1:1 mol HCl required = mol = mol/l V V = L 157

14 Check Your Answer Since these reagents react in a mol ratio of 1:1, compare the amount of each using rounded off values: amount HCl = = mol; mol NaOH = = The values are about the same and therefore the answer is reasonable. 17. Problem Sodium sulfate and water result when sulfuric acid, H 2 SO 4, reacts with sodium hydroxide. What volume of mol/l sulfuric acid is needed to react with 47.3 ml of mol/l aqueous sodium hydroxide? You must find the volume of H 2 SO 4 that reacts with the NaOH. You know a volume and concentration of NaOH and a concentration of HCl. Calculate the mol of NaOH using n = C V. Write the balanced equation for this reaction. Use the mol ratio in the balanced equation to calculate the mol of H 2 SO 4 that combines with this number of mol of NaOH. n Use the formula V = to calculate the volume of H 2 SO 4. C H 2 SO 4(aq) + 2 NaOH (aq) Æ Na 2 SO 4(aq) + 2 H 2 O (l) mol NaOH = mol/l L = mol mol NaOH 1 mol H2SO4 = mol H 2 SO 4 2 mol NaOH mol H 2 SO 4 = mol/l V V = L Check Your Answer Since these reagents react in a mol ratio of 1:2, compare the number of mol of each using rounded off values. mol NaOH = = 0.11; mol H 2 SO 4 = = 0.06 The values are about in a ratio of 1:2 and therefore our answer is reasonable. 18. Problem Your stomach secretes hydrochloric acid to help you digest the food you have eaten. If too much HCl is secreted, however, you may need to take an antacid to neutralize the excess. One antacid product contains the compound magnesium hydroxide, Mg(OH) 2. (a) Predict the reaction that takes place when magnesium hydroxide reacts with hydrochloric acid (Hint: this is a double displacement reaction) (b) Imagine that you are a chemical analyst testing the effectiveness of antacids. If 0.10 mol/l HCl serves as your model for stomach acid, how many litres will react with an antacid that contains 0.10 g of magnesium hydroxide. (a) Write the equation for the double displacement reaction between hydrochloric acid and magnesium hydroxide. (b) Find the volume of 0.10 mol/l HCl that will react with 0.10 g Mg(OH) 2 You are given a mass of Mg(OH) 2, a concentration of HCl and the periodic table to find the molar mass of Mg(OH) 2 158

15 Write the balanced equation for this double displacement reaction. Calculate the m molar mass of Mg(OH) 2 and convert 0.10 g to moles using n =. Use the M mol ratio in the balanced equation to determine the mol of HCl that will react. n Determine the volume of HCl using V =. C (a) Mg(OH) 2(aq) + 2 HCl (aq) Æ MgCl 2(aq) + 2H 2 O (l) (b) M for Mg(OH) 2 = g/mol 0.10 g mol Mg(OH) 2 = = mol g/mol mol Mg(OH) 2 2 mol HCl = mol HCl mol 1 mol Mg()H) 2 V = = L 0.10 mol/l Therefore the volume of HCl that reacts is 34 ml. Working backwards, using rounded off numbers, and a mol ratio of 1:2, L 0.10 mol/l = mol. This is about the mol of Mg(OH) 2 that have been given. The answer 2 is reasonable. Solutions for Practice Problems Student Textbook pages Problem 8.76 g of sodium sulfide is added to 350 ml of mol/l lead(ii) nitrate. Calculate the maximum mass of precipitate that can form. What Is Required? You must calculate the mass of precipitate that forms when known amounts of solutions Na 2 S and Pb(NO 3 ) 2 react. What Is Given? You know a mass of Na 2 S and a volume and concentration of Pb(NO 3 ) 2 solution. Find the molar mass of Na 2 S and calculate the moles of this compound. Use the formula n = C V to calculate the moles of Pb(NO 3 ) 2. Write the balanced equation for the reaction between these two compounds and refer to the solubility guideline to identify the insoluble product. Identify which of these two reactants is the limiting reagent. Use the limiting reagent to calculate the moles of precipitate. Determine the molar mass of the precipitate and convert the moles to grams. molar mass of Na 2 S = g/mol n = = = mol Na 2 S mass molar mass 8.76 g g/mol n = C V = mol/l L = mol Pb(NO 3 ) 2 Na 2 S (aq) + Pb(NO 3 ) 2(aq) 2NaNO 3(aq) + PbS (s) The precipitate is PbS. Since the reactants react in a molar ratio of 1:1, the compound with the lesser number of moles will be the limiting reagent, i.e. the mol Pb(NO 3 ) 2. Also, since the mol ratio of Pb(NO 3 ) 2 : PbS is 1:1, we can conclude that mol of Pb(NO 3 ) 2 will produce mol of PbS. 159

16 molar mass of PbS = g/mol mass of PbS mol g 1mol = 20.9 g of PbS The final answer has the correct unit and number of significant figures. This answer seems reasonable based upon the mass of the sample. 20. Problem 25.0 ml of mol/l Pb(NO 3 ) 2(aq) is mixed with 300 ml of mol/l KI (aq). What is the maximum mass of precipitate that can form? What Is Required? Find the mass of precipitate that will form when known volumes of two solutions of given concentration are mixed? What Is Given? You are given the volumes, concentrations, names and formulae of two solutions that are to be mixed. Write the balanced equation for this reaction and identify the insoluble product. Use the formula n = C V to calculate the number of moles of each reactant. Identify the limiting reagent. Use the limiting reagent to calculate the moles of precipitate. Find the molar mass of the precipitate and convert the moles to grams. Pb(NO 3 ) 2(aq) + 2KI (aq) 2KNO 3(aq) + PbI 2(s) moles Pb(NO 3 ) 2(aq) = n = C V = mol/l L = mol moles KI (aq) = n = C V = mol/l L = mol From the balanced equation, Pb(NO 3 ) 2 : KI = 1 : mol Pb(NO 3 ) 2molKI 1mol Pb(NO 3 ) 2 = mol KI But there are mol of KI available, therefore KI is in excess and Pb(NO 3 ) 2 is the limiting reagent. From the balanced equation, the molar ratio Pb(NO 3 ) 2 : PbI 2 is 1:1. Therefore, mol of PbI2 are produced. molar mass of PbI 2 = g/mol mass of PbI 2 = mol g mol = 4.61 g PbI 2 The final answer has the correct unit and number of significant figures. This answer seems reasonable based upon the mass of the sample. 21. Problem Zoe mixes 15.0 ml of mol/l aqueous sodium hydroxide with 20.0 ml of mol/l aqueous aluminum nitrate. (a) Write the chemical equation for the reaction. (b) Calculate the maximum mass of precipitate that forms. What is required? You must write a balanced equation for this reaction and calculate the mass of precipitate that forms. You have a volume and concentration for each of the reactants and the periodic table to determine molar masses. 160

17 This reaction is a double displacement reaction. Refer to the solubility guidelines in the textbook to determine which product is the precipitate. Write the balanced equation. Calculate the mol of NaOH and Al(NO 3 ) 2 using n = C V and use the mole ratio in the balanced equation to determine the limiting reagent. Use mol of limiting reagent and the mol ratio in the balanced equation to calculate the mol of precipitate, Al(OH) 3. Use the formula m = n M to calculate the mass of Al(OH) 3. (a) 3 NaOH (aq) + Al(NO 3 ) 2(aq) Æ 3 NaNO 3(aq) + Al(OH) 3(s) (b) mol NaOH (aq) = mol/l L = mol mol Al(NO 3 ) 2(aq) = mol/l L = mol NaOH (aq) and Al(NO 3 ) 2(aq) react in a ratio of 3: mol NaOH 1 mol Al(NO3 ) 2 = mol Al(NO 3 ) 2(aq) 3 mol NaOH Since mol of Al(NO 3 ) 2(aq) are given, and mol are required to react with all of the NaOH, Al(NO 3 ) 2(aq) is in excess and NaOH is the limiting reagent. 1 mol Al(OH) mol NaOH = mol Al(OH) 3 mol NaOH 3(s) molar mass Al(OH) 3 = g/mol mass of Al(OH) 3 = mol g/mol = g Therefore the mass of precipitate, Al(OH) 3, is g Working backwards from the mass of precipitate, verify that the mol of limiting reagent corresponds to the data given in the question. the answer has the correct unit and number of significant figures and seems reasonable. 22. Problem In an experiment, Kendra mixed 40.0 ml of mol/l lead(ii) nitrate with 50.0 ml of 1.22 mol/l hydrochloric acid. A white precipitate formed which Kendra filtered and dried. She determined the mass to be g. (a) What is the formula of the precipitate that formed? (b) Which is the limiting reactant? (c) What is the theoretical yield? (d) Calculate the percentage yield. You must determine what compound precipitates and write its formula. Write a balanced equation for the reaction and determine which is the limiting reactant. Finally, you must calculate the theoretical and percentage yields. You know the volume and concentration of each reactant and can use the periodic table to determine molar masses. (a) This reaction is a double displacement reaction. Refer to the solubility guidelines in the student textbook to determine which product is the precipitate. (b) Write the balanced equation. (c) Calculate the mol of Pb(NO 3 ) 2(aq) and HCl (aq) using n = C V and use the mole ratio in the balanced equation to determine the limiting reagent. Use mol of limiting reagent and the mol ratio in the balanced equation to calculate the mol of 161

18 precipitate, PbCl 2(s) Use the formula m = n M to calculate the mass of PbCl 2(s). This is the theoretical yield. (d) Calculate the percentage yield using the formula actual yield percentage yield = 100% theoretical yield (a) The precipitate is PbCl 2 (b) 2 HCl (aq) + Pb(NO 3 ) 2(aq) Æ 2 HNO 3(aq) + PbCl 2(s) (c) mol HCl (aq) = 1.22 mol/l L = mol mol Pb(NO 3 ) 2(aq) = mol/l L = mol mol HCl (aq) 1 mol PbCl2 = mol PbCl 2(s) 1 mol Pb(NO mol Pb(NO 3 ) 2(aq) 1 mol PbCl2 = mol PbCl 2(s) 1 mol Pb(NO 3) 2 Since the Pb(NO 3 ) 2(aq) gives a smaller amount of PbCl 2(s), Pb(NO 3 ) 2(aq) is the limiting reagent. molar mass of PbCl 2(s) = g/mol mass of PbCl 2(s) = mol g/mol = 6.15 g PbCl 2(s) Therefore the theoretical yield of PbCl 2(s) is 6.15 g. actual yield (d) percentage yield = theoretical yield 100% = g 100% = 81.5 % 6.15 g Therefore the percentage yield is 81.5% 3 ) 2 Work backwards to verify that this yield matches the data in the question. The answer seems reasonable and has the correct number of significant figures and units. 162