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Get Full Access to Introductory Chemistry - 5 Edition - Chapter 14 - Problem 108p
Get Full Access to Introductory Chemistry - 5 Edition - Chapter 14 - Problem 108p

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# Solved: A popular classroom demonstration consists of filing notches into a new penny

ISBN: 9780321910295 34

## Solution for problem 108P Chapter 14

Introductory Chemistry | 5th Edition

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Problem 108P

A popular classroom demonstration consists of filing notches into a new penny and soaking the penny in hydrochloric acid overnight. Because new pennies are made of zinc coated with copper, and hydrochloric acid dissolves zinc and not copper, the inside of the penny is dissolved by the acid, while the outer copper shell remains. Suppose a penny contains 2.5 g of zinc and is soaked in 20.0 mL of 6.0 M HCl. Calculate the concentration of the HCl solution after all of the zinc has dissolved. Hint: The Zn from the penny is oxidized to Zn2+

Step-by-Step Solution:

Solution 108P :

Step 1:

Given :

Mass of zinc in penny = 2.5 g

Volume of HCl = 20.0 mL

Molarity of HCl = 6.0 M

Concentration of HCl solution after all of the zinc has dissolved = ?

First, let’s write the chemical equation for the given scenario :

Zn(s) + 2HCl(aq) →  ZnCl2(aq) + H2(g)

Now, let’s calculate the moles of Zinc :

Number of moles (n) =

Molar mass of Zinc(Zn)  = 65.38 g/mol

Therefore, n =

= 0.0382 moles

Now, let’s calculate the moles of HCl:

We have the molarity and volume of HCl , using this we will calculate the number of moles using the molarity formula :

Molarity =

The formula for calculating the number of moles will be :

Number of moles = Molarity  Volume(in L)

Volume of HCl is in mL, let’s convert to L :

1 L = 1000 mL

Therefore, 20 mL in L will be:

= 20 mL

= 0.020 L

Now, let’s calculate the number of moles :

= 6.0 M 0.020 L

= 0.12 mol

Step 2 of 3

Step 3 of 3

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