Types of Chemical Reactions and Solution Stoichiometry Chapter 4 Types of Chemical Reactions and Solution Stoichiometry

QUESTION Which statement below best describes the connection between water and electrolytes? 1. Polar water molecules enable weak electrolytes to dissociate completely to brightly light a conductivity apparatus. 2. Strong electrolytes do not need to become attached to polar water molecules to conduct electricity. 3. Non-electrolytes do not interact with polar water. 4. In strong electrolytes conductivity can arise from dissociation or ionization. HMClass Prep: Figure 4.2 Copyright © Houghton Mifflin Company. All rights reserved.

ANSWER Choice 4 is correct. Strong acids such as HCl are not ionic until water interacts with HCl molecules to facilitate ionization. Ionic salts dissociate into separate ions. Section 4.2: The Nature of Aqueous Solutions: Strong and Weak Electrolytes Copyright © Houghton Mifflin Company. All rights reserved.

QUESTION If 250.0 mL of 0.200 M NaCl were evaporated without the loss of any solute, how many grams of NaCl would remain? 1. 2.92 g 2. 11.6 g 3. 2 900 g 4. I do not know how to use these values to determine the answer. Copyright © Houghton Mifflin Company. All rights reserved.

ANSWER Choice 1 provides the correct number of grams. When volume, in liters, is multiplied by molarity, moles of solute can be determined. Once this is known it can be converted to grams using the solute’s molar mass. Section 4.3: The Composition of Solutions Copyright © Houghton Mifflin Company. All rights reserved.

QUESTION If an antacid contains Al(OH)3 it will form AlCl3 upon neutralization of stomach acid. What would be the molarity of Cl– ions in 100.0 mL of 0.010 M AlCl3? 1. 0.001 0 M 2. 0.010 M 3. 0.003 0 M 4. 0.030 M Copyright © Houghton Mifflin Company. All rights reserved.

ANSWER Choice 3 accounts for both the stoichiometry of dissociation and the concentration. After multiplying the volume, in liters, by the molarity of the solution, the stoichiometry must be considered. AlCl3 dissociates into 3 moles of Cl–. Section 4.3: The Composition of Solutions Copyright © Houghton Mifflin Company. All rights reserved.

QUESTION The acid in car batteries is sulfuric acid, H2SO4. Besides water, what reaction product would likely form when sulfuric acid reacts with calcium hydroxide? Would the product likely be a strong or weak electrolyte? 1. Ca(SO4)2 weak 2. CaSO4 weak 3. CaSO4 strong 4. Ca(SO4)2 strong Copyright © Houghton Mifflin Company. All rights reserved.

ANSWER Choice 2 indicates the correct formula and rating for electrolyte behavior. The anion of the acid reacts with the cation of the base. Ca has a +2 ionic charge so one sulfate ion (with –2 charge) is needed to balance the calcium charge. Section 4.2: The Nature of Aqueous Solutions: Strong and Weak Electrolytes Section 4.5: Precipitation Reactions Copyright © Houghton Mifflin Company. All rights reserved.

QUESTION A 0.14 M solution of NaCl provides the same concentration of NaCl typically found in blood serum. If you had a 0.50 M solution of NaCl but wanted to produce 200.0 mL of 0.14 M NaCl, how many mL of the concentrated solution would be needed to make the dilution? 1. 28 mL 2. 56 mL 3. 14 mL 4. 200 mL Copyright © Houghton Mifflin Company. All rights reserved.

ANSWER Choice 2 provides the correct number of mL of the more concentrated solution. The first step is to determine the number of moles of NaCl needed in the final solution (0.200 0 L  0.14 mol/L = 0.028 moles), then determine the volume of the concentrated solution that contains that many moles. (0.028 moles = V  0.50 mol/L) Section 4.3: The Composition of Solutions Copyright © Houghton Mifflin Company. All rights reserved.

QUESTION HMClass Prep: Table 4.1 Copyright © Houghton Mifflin Company. All rights reserved.

QUESTION (continued) Of the following compounds which, when placed in distilled water, would give nearly the same conductivity result as pure water alone? 1. CuSO4 2. (NH4)2S 3. AgCl 4. FeCl3 Copyright © Houghton Mifflin Company. All rights reserved.

ANSWER Choice 3 represents an insoluble compound. The proper use of the solubility rules chart reveals that most chlorides dissociate in water. AgCl does not appreciably do so. Section 4.5: Precipitation Reactions Copyright © Houghton Mifflin Company. All rights reserved.

QUESTION HMClass Prep: Table 4.1 Copyright © Houghton Mifflin Company. All rights reserved.

QUESTION (continued) If you began a reaction with the following ions in solution (all would be written with an aq subscript) how would you represent the proper final net ionic equation? 6Na+ + 2PO43– + 3Fe2+ + 6NO3–  1. 3Na+ + PO43– + Fe2+ + 2NO3–  No Reaction 2. 6Na+ + 2PO43– + 3Fe2+ + 6NO3–  Fe3(PO4)2 (s)+ 6NaNO3 3. 3Na+ + PO43– + Fe2+ + 2NO3–  Fe3(PO4)2 (s)+ 6 Na+ + 6 NO3– 4. 2PO43– + 3Fe2+  Fe3(PO4)2 (s) Copyright © Houghton Mifflin Company. All rights reserved.

ANSWER Choice 4 correctly depicts the reacting ions and the resulting solid product. Use the chart to determine if an insoluble solid will be produced by the combination of ions given. If so, write that and balance the atoms. If soluble ions from strong electrolytes are present, they may cancel out from the left and right side of the equation. Section 4.5: Precipitation Reactions Copyright © Houghton Mifflin Company. All rights reserved.

__ _____ + ____ _____  Al2(CO3)3 QUESTION Given the insoluble compound Al2(CO3)3 predict the ions and coefficients that would be necessary to complete the following net ionic equation: __ _____ + ____ _____  Al2(CO3)3 1. 2 AlCl3 + 3 Na2CO3 also include 6 NaCl on right 2. 3 Al3+ + 2 CO32– 3. 2 Al3+ + 3 CO32– 4. 2 Al3+ 6 Cl – + 3 CO32– + 6 Na+ Copyright © Houghton Mifflin Company. All rights reserved.

ANSWER Choice 3 correctly depicts the completed net ionic equation. Only those soluble reacting ions that form the compound need to be shown. The other ions could be considered spectator ions. Section 4.6: Describing Reactions in Solution Copyright © Houghton Mifflin Company. All rights reserved.

QUESTION If ammonium carbonate and phosphoric acid are mixed what would you expect to form? 1. Ammonium phosphate + hydrogen carbonate 2. Ammonium phosphate + water 3. Ammonium phosphate + hydrogen + water 4. Ammonium phosphate + water + carbon dioxide Copyright © Houghton Mifflin Company. All rights reserved.

ANSWER Choice 4 properly indicates that when carbonates react with acids, carbon dioxide gas is produced (along with water and a salt). Section 4.6: Describing Reactions in Solution Copyright © Houghton Mifflin Company. All rights reserved.

QUESTION Suppose a 250.0 mL water sample has a lead nitrate Pb(NO3)2 concentration of 0.10 M. If you wanted to precipitate all the lead from the water you could add a solution containing NaCl to get PbCl2 to precipitate; thus removing the lead. How many mL of 0.50 M NaCl would you have to add to your sample to precipitate all the Pb2+? 1. 50. mL 2. 1.0  102 mL 3. 2.0  102 mL 4. I know this should be possible, but I do not get any of those answers. HMClass Prep: Figure 4.17b Copyright © Houghton Mifflin Company. All rights reserved.

ANSWER Choice 2 provides the consideration of stoichiometry and solution concentrations. First, consider that multiplying the volume  M of lead nitrate will provide the number of moles of Pb2+ that need to be precipitated. Next, note that each lead ion will need two chloride anions. To keep that 1:2 mole ratio, twice as many moles of Cl– than Pb2+ are required. Once the Cl– anions are known, then the volume containing that number of moles can be calculated. Moles = V  M Section 4.7: Stoichiometry of Precipitation Reactions Copyright © Houghton Mifflin Company. All rights reserved.

QUESTION In order to facilitate x-ray examinations of some internal organs, patients drink a solution containing nearly insoluble BaSO4. Good thing this is insoluble, otherwise we would be exposed to toxic barium ions! How many moles of BaSO4 could be precipitated from mixing 250.0 mL of 0.20 M Ba(OH)2 with 125 mL of 0.10 M H2SO4? What is the limiting reagent? 1. 0.050 moles; barium hydroxide is the limiting reagent. 2. 0.050 moles; sulfuric acid is the limiting reagent. 3. 0.012 moles; barium hydroxide is the limiting reagent. 4. 0.012 moles; sulfuric acid is the limiting reagent. HMClass Prep: Figure 4.17c Copyright © Houghton Mifflin Company. All rights reserved.

ANSWER Choice 4 properly shows both the number of moles and identifies the limiting reagent. Volume (in liters)  M = moles. This needs to be done for both reagents. Then the smallest can be used to predict the maximum yield of product (because this reaction has 1:1 stochiometry). Section 4.7: Stoichiometry of Precipitation Reactions Copyright © Houghton Mifflin Company. All rights reserved.

QUESTION Phosphoric acid is often found in soft drinks. How many mL of 0.100 M NaOH would be required to neutralize 300.0 mL of 0.001 0 M phosphoric acid? 1. 30. mL 2. 3.0  103 mL 3. 9.0 mL 4. 10. mL HMClass Prep: Figure 4.17c Copyright © Houghton Mifflin Company. All rights reserved.

ANSWER Choice 3 properly considers the concentrations and stoichiometry. Phosphoric acid is H3PO4 so each mole requires 3 moles of NaOH to neutralize. For phosphoric: V  M = moles of phosphoric initially. Three times this shows how many moles of NaOH are needed. Then the moles of NaOH can be used to determine the volume needed from a 0.10 M solution. Section 4.8: Acid–Base Reactions Copyright © Houghton Mifflin Company. All rights reserved.

Au(s) + NO3–(aq) + Cl–(aq)  AuCl4–(aq) + NO2(g) QUESTION One reason gold is so prized is because it seldom reacts. However, a mixture of nitric acid and hydrochloric acid (called aqua regia; i.e. king water) will, through a redox reaction, dissolve gold. When the following redox equation is properly balanced, how many H+ ions will appear on the left side of the equation? Au(s) + NO3–(aq) + Cl–(aq)  AuCl4–(aq) + NO2(g) 1. 0 2. 2 3. 4 4. 6 Copyright © Houghton Mifflin Company. All rights reserved.

ANSWER Choice 4 reflects the correct H+ ion count for the balanced equation. Section 4.10: Balancing Oxidation–Reduction Equations Copyright © Houghton Mifflin Company. All rights reserved.

MnO4– + C2O42–  MnO2 + CO32– (basic solution) QUESTION Oxalate ion can be found in rhubarb and spinach (among other green leafy plants). The following unbalanced equation carried out in a basic solution, shows how MnO4– could be used to analyze samples for oxalate. MnO4– + C2O42–  MnO2 + CO32– (basic solution) When properly balanced, how many OH– are present? 1. 1 2. 2 3. 3 4. 4 Copyright © Houghton Mifflin Company. All rights reserved.

ANSWER Choice 4 is correct. The redox equation could be balanced according to the steps for an acid solution, but then OH– must be added to neutralize the H+ ions. When done properly 4 OH– ions will be present in the basic equation. Section 4.10: Balancing Oxidation–Reduction Equations Copyright © Houghton Mifflin Company. All rights reserved.