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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,...

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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

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