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What is the minimum amount of 6.0 M H2SO4 necessary to

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

Solution for problem 67E 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 67E

Problem 67E

What is the minimum amount of 6.0 M H2SO4 necessary to produce 25.0 g of H2 (g) according to the reaction between aluminum and sulfuric acid?

Step-by-Step Solution:
Step 1 of 3

Solution: Here, we are going to calculate the mass of ethanol required to make 1.00 L of vodka. Also, we are going to calculate the volume of ethanol required to make 1.00 L of vodka. Step1: Molarity is defined as the number of moles of the solute(Here ethanol is the solute) in 1 litre of the solution. Mathematically, Number of moles of solute Molarity(M) = -------------------------------------- ------(1) Volume of solution in litres Step2: a) Given, molarity of the ethanol solution = 6.86 M Volume of the ethanol solution in Litres = 1.00 L Using the mathematical expression for molarity calculation, we get, number of moles of ethanol = Volume of solution in L X Molarity ethanol = 1.00 l X 6.86 M = 6.86 mol. Now, 1 mole of ethanol(CH CH3H) 2molar mass of ethanol Therefore, 6.86 mol of ethanol = 6.86 X molar mass of ethanol = 6.86 X 46.0682 g [molar mass of CH 3 2 = 46.0682 g] = 316.028 g Thus, 316.028 grams of ethanol should be dissolved in water to make 1.00 L of vodka. Step3: b) Density can be defined as mass per unit volume. i.e., Density = mass / volume Therefore, volume = mass / density Here, mass of ethanol = 316.028 g Density of ethanol = 0.789 g/mL Therefore, volume of ethanol = 316.028 g / 0.789 (g/mL) = 400.54 mL Thus, 400.54 mL of ethanol is required to make 1.00 L of vodka. ------------------

Step 2 of 3

Chapter 4, Problem 67E is Solved
Step 3 of 3

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

The answer to “What is the minimum amount of 6.0 M H2SO4 necessary to produce 25.0 g of H2 (g) according to the reaction between aluminum and sulfuric acid?” is broken down into a number of easy to follow steps, and 26 words. The full step-by-step solution to problem: 67E from chapter: 4 was answered by , our top Chemistry solution expert on 02/22/17, 04:35PM. This textbook survival guide was created for the textbook: Chemistry: A Molecular Approach, edition: 3. Chemistry: A Molecular Approach was written by and is associated to the ISBN: 9780321809247. Since the solution to 67E from 4 chapter was answered, more than 366 students have viewed the full step-by-step answer. This full solution covers the following key subjects: ethanol, vodka, make, need, dissolve. This expansive textbook survival guide covers 82 chapters, and 9454 solutions.

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What is the minimum amount of 6.0 M H2SO4 necessary to