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: Consider the following reagents: zinc, copper, mercury

Chemistry: A Molecular Approach | 3rd Edition | ISBN: 9780321809247 | Authors: Nivaldo J. Tro ISBN: 9780321809247 1

Solution for problem 97AE Chapter 4

Chemistry: A Molecular Approach | 3rd Edition

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Chemistry: A Molecular Approach | 3rd Edition | ISBN: 9780321809247 | Authors: Nivaldo J. Tro

Chemistry: A Molecular Approach | 3rd Edition

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Problem 97AE

97AE: Consider the following reagents: zinc, copper, mercury (density 13.6 g/mL), silver nitrate solution, nitric acid solution, (a) Given a 500-mL Erlenmeyer flask and a balloon, can you combine two or more of the foregoing reagents to initiate a chemical reaction that will inflate the balloon? Write a balanced chemical equation to represent this process. What is the identity of the substance that inflates the balloon? (b) What is the theoretical yield of the substance that fills the balloon? (c) Can you combine two or more of the foregoing reagents to initiate a chemical reaction that will produce metallic silver? Write a balanced chemical equation to represent this process. What ions are left behind in solution? (d) What is the theoretical yield of silver?

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Step 1 of 3

Solution: Step1: We can say that all metals are not equally reactive. That is why, reactive metals can displace less reactive metals from their compounds in solution or molten form. In order to know, which metal is more reactive, several experiments were performed, and finally a series known as the reactivity or activity series was developed.The metals at the top of the activity series are readily oxidized whereas those at the bottom of the activity series are least readily oxidized. Step2: a) In order to inflate a balloon, hydrogen gas must be produced. Among all the reagents given above, Zinc is more reactive than hydrogen as it is placed above hydrogen in the reactivity series. Thus, Zinc will liberate hydrogen gas from nitric acid when combined. The process can be represented by the following chemical equation: Zn(s) + 2HNO (a3 --------> Zn(NO ) (3 2 H (g2 ---(1) The hydrogen gas produced in the reaction inflates the balloon. Step3: b) Given, mass of zinc = 35.0 g Molar mass of zinc = 65.38 g/mol 35 Therefore, number of moles of zinc = 65.38 = 0.535 mol Again, Volume of nitric acid solution used = 150 mL = 1000 L = 0.150 L Molarity of nitric acid solution = 3.00 M Therefore, number of moles of nitric acid = Volume in litres X Molarity = 0.150 L X 3.00 M = 0.450 mol Step4: Now, from equation (1), it is seen that 2 mol of nitric acid reacts with 1 mol of zinc. Therefore, 0.450 mol of nitric acid will react with ( x 2.450 = 0.225) mol of zinc. Step5: Again, 1 mol of zinc produces 1 mol of hydrogen gas(H ). 2 Therefore, 0.225 mol of zinc will produce 0.225 mol of hydrogen gas. Now, 1 mol of hydrogen gas = molar mass of hydrogen gas 0.225 mol of hydrogen gas = 0.225 x molar mass of hydrogen gas = 0.225 x 2 x atomic mass of hydrogen = 0.450 x 1.0079 = 0.453 g Thus, 0.453 g of hydrogen gas is produced. Step6: c) Among all the reagents given above, Cu is more reactive than silver as it is placed above silver in the reactivity series. Thus, Copper will replace silver from its nitrate solution. The process can be represented by the following chemical equation: Cu(s) + 2AgNO (aq3--------> Cu(NO ) (a3 22Ag(s) ---(2) Since, out of the two products, copper nitrate is soluble in water, therefore, the ions left behind 2+ - are Cu and NO . 3 Step7: d) Given, mass of copper = 42.0 g Molar mass of copper = 63.546 g/mol Therefore, number of moles of copper = 63.546 = 0.661 mol 150 Again, Volume of silver nitrate solution used = 150 mL = 1000L = 0.150 L Molarity of silver nitrate solution = 0.750 M Therefore, number of moles of silver nitrate = Volume in litres X Molarity = 0.150 L X 0.750 M = 0.1125 mol Step8: Now, from equation (2), it is seen that 2 mol of silver nitrate reacts with 1 mol of copper. 1 Therefore, 0.1125 mol of silver nitrate will react with ( x 021125 = 0.05625) mol of copper. Step9: Again, 1 mol of copper produces 2 mol of silver metal(Ag). Therefore, 0.05625 mol of copper will produce (0.05625 x 2 = 0.1125) mol of silver metal. Now, 1 mol of silver metal = molar mass of silver 0.1125 mol of silver = 0.1125 x molar mass of silver = 0.1125 x 107.8682 g [molar mass of Ag = 107.8682 g/mol] = 12.135 g Thus, 12.135 g of hydrogen gas is produced. ----------------

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Chapter 4, Problem 97AE is Solved
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Textbook: Chemistry: A Molecular Approach
Edition: 3
Author: Nivaldo J. Tro
ISBN: 9780321809247

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: Consider the following reagents: zinc, copper, mercury