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One drop (assume 0.050 mL) of 6.0 M HCl is placed onto the

Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro ISBN: 9780321910295 34

Solution for problem 107P Chapter 16

Introductory Chemistry | 5th Edition

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Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro

Introductory Chemistry | 5th Edition

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Problem 107P

One drop (assume 0.050 mL) of 6.0 M HCl is placed onto the surface of 0.028-mm-thick aluminum foil. What is the maximum diameter of the hole that will result from the HCl dissolving the aluminum? (Density of aluminum = 2.7 g/cm3)

Step-by-Step Solution:
Step 1 of 3

Solution 107P:

Here, we are going to calculate the diameter of the hole that will result from the HCl dissolving the aluminium.

Step1:

Given,

Volume of HCl = 0.050 mL = 0.050 / 1000 L = 0.000050 L                [1000 mL = 1 L]

Molarity of HCl = 6.0 M

Therefore, number of moles of HCl = Volume of HCl x Molarity

                                        = 0.000050 L x 6.0 M

                                        = 0.0003 mol

Step2:

The reaction between Al and HCl takes place in the following manner:

                                2Al + 6HCl → 2AlCl3 + 3H2

From the above equation, it is clear that, 6 mol of HCl dissolves 2 mol of Al.

Therefore, 0.0003 mol of HCl will dissolve (2/6 x 0.0003 = 0.0001) mol of Al

Step3:

1 mol Al =  molar mass of Al

Therefore, 0.0001 mol Al = 0.0001 x molar mass of Al

                              = 0.0001 x 26.98 g/mol

                              = 0.002698 g

Step4:

Density of Al = 2.7 g/cm3

Therefore, volume of Al reacted = mass of Al / density

                                    = 0.002698 g / 2.7 g cm-3

                                    = 0.001 cm3

Step5:

Since the hole is cylindrical in shape, therefore, volume of the hole is given by = 𝛑r2h, where, r is the radius of the hole and h is the thickness.

Volume of the hole = volume of Al reacted = 0.001 cm3

Thickness of the hole = thickness of the Al foil = 0.028 mm

    = 0.028 / 10 cm         [10 mm = 1 cm]

    = 0.0028 cm

Substituting the values in the formula, we have,

3.14 x r2 x 0.0028 cm = 0.001 cm3

      0.008792 cm x r2 = 0.001 cm3

                      r2 = 0.001 cm3 / 0.008792 cm

                      r2 = 0.11 cm

                      r =

                      r = 0.33 cm (approx.)

Therefore, diameter = 2 x radius

                        = 2 x 0.33 cm

                        = 0.66 cm

Thus, the diameter of the hole is 0.66 cm.

                                        ------------------------

Step 2 of 3

Chapter 16, Problem 107P is Solved
Step 3 of 3

Textbook: Introductory Chemistry
Edition: 5
Author: Nivaldo J Tro
ISBN: 9780321910295

This textbook survival guide was created for the textbook: Introductory Chemistry, edition: 5. The answer to “One drop (assume 0.050 mL) of 6.0 M HCl is placed onto the surface of 0.028-mm-thick aluminum foil. What is the maximum diameter of the hole that will result from the HCl dissolving the aluminum? (Density of aluminum = 2.7 g/cm3)” is broken down into a number of easy to follow steps, and 41 words. Introductory Chemistry was written by and is associated to the ISBN: 9780321910295. This full solution covers the following key subjects: Aluminum, HCL, Density, diameter, dissolving. This expansive textbook survival guide covers 19 chapters, and 2046 solutions. Since the solution to 107P from 16 chapter was answered, more than 409 students have viewed the full step-by-step answer. The full step-by-step solution to problem: 107P from chapter: 16 was answered by , our top Chemistry solution expert on 05/06/17, 06:45PM.

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