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IE: A sample of 7.75 g of Mg(OH)2 is added to 25.0 ml of

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

Solution for problem 109IE 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 109IE

IE: A sample of 7.75 g of Mg(OH)2 is added to 25.0 ml of 0.200 M HN03. (a) Write the chemical equation for the reaction that occurs, (b) Which is the limiting reactant in the reaction? (c) How many moles of Mg(OH)2, HN03, and Mg(N03)2 are present after the reaction is complete?

Step-by-Step Solution:
Step 1 of 3

Solution 109 IE: Step1: a) Magnesium hydroxide reacts with nitric acid to form magnesium nitrate and water. The chemical equation for the reaction is: Mg(OH) (aq2+ 2HNO (aq) --3--> Mg(NO ) (aq) + 2H3 2) 2 Step2: b) Given, mass of Mg(OH) added = 725 g Molar mass of Mg(OH) = 58.322/mol Therefore, number of moles of Mg(OH) = mass of Mg(OH) / molar mass 2 7.75 2 = 58.32 = 0.133 mol Again, molarity of the HNO soluti3 = 0.200 M Volume of the HNO solution = 25.0 mL = 25 L = 0.025 L 3 1000 Therefore, number of moles of HNO solution =3olume of solution in litres X Molarity = 0.025 L X 0.200 M = 0.005 mol. Step3: From the equation above, it is clear that, 2 mol of HNO reacts with3 mol of Mg(OH) 2 for...

Step 2 of 3

Chapter 4, Problem 109IE is Solved
Step 3 of 3

Textbook: Chemistry: A Molecular Approach
Edition: 3
Author: Nivaldo J. Tro
ISBN: 9780321809247

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