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Get Full Access to Chemistry: A Molecular Approach - 3 Edition - Chapter 4 - Problem 84e
Get Full Access to Chemistry: A Molecular Approach - 3 Edition - Chapter 4 - Problem 84e

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# Solved: Write balanced molecular and net ionic equations ISBN: 9780321809247 1

## Solution for problem 84E Chapter 4

Chemistry: A Molecular Approach | 3rd Edition

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Problem 84E

Problem 84E

Write balanced molecular and net ionic equations for the reaction between nitric acid and calcium hydroxide.

Step-by-Step Solution:
Step 1 of 3

Solution: Here, we are going to calculate the mass of acetic acid that must be added to neutralize 42.5 mL of 0.115 M sodium hydroxide. Step1: Given, volume of NaOH solution used = 42.5 mL = 42.5/1000 L = 0.0425 L Molarity of NaOH solution used = 0.115 M Therefore, number of moles of NaOH = Volume of solution in litres X Molarity = 0.0425 L X 0.115 M = 0.005 mol Step2: Acetic acid reacts with sodium hydroxide according to the following way: CH C3H(aq) + NaOH(aq) ------> NaCH COO(aq) + H O 3 2 In the above equation, 1 mole of NaOH requires 1 mole of CH COOH for complete 3 neutralization. Therefore, 0.005 moles of NaOH will require 0.005 moles of CH COOH for complete 3 neutralization. Step3: Now, 1 qt = 946.353 mL 3.45 mL of CH COOH 3 equivalent to 0.005 moles of CH COOH 3 Therefore, 946.353 mL of CH COOH will 3 equivalent to [(0.005/3.45) X 946.353] mol of CH 3OH i.e., 946.353 mL of CH COOH will be3quivalent to 1.37 mole of CH COOH. 3 Step4: Now, 1 mole of CH COOH =3olar mass of CH COOH 3 Therefore, 1.37 moles of CH COOH = 137 X molar mass of CH COOH 3 = 1.37 X 60.05 g [ molar mass of CH COOH = 60.05 3 g/mol] = 82.3 g Thus, the required mass of CH COOH is 3.3 g.

Step 2 of 3

Step 3 of 3

##### ISBN: 9780321809247

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