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# The decomposition of a silver oxide sample forms 15.8 g of Ag(s):2 Ag2O(s) ? 4Ag(s) + ISBN: 9780321910295 34

## Solution for problem 109P Chapter 11

Introductory Chemistry | 5th Edition

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Problem 109P

PROBLEM 109P

The decomposition of a silver oxide sample forms 15.8 g of Ag(s):

2 Ag2O(s) → 4Ag(s) + O2(g)

What total volume of O2 gas forms if it is collected over water at a temperature of 25 °C and a total pressure of 752 mm Hg?

Step-by-Step Solution:
Step 1 of 3

Solution 109P

Given that

Your reaction will produce 0.94 L of oxygen gas.

The first thing you need to do is determine how many moles of oxygen gas are produced when that much silver is produced. To do this, use the number of moles of silver and the mole ratio that exists between the two products.

2Ag2O(s)→4Ag(s)+O2(g)

Notice that you have a 4:1 mole ratio between silver and oxygen, which means that you get 1 mole of oxygen for every 4 moles of silver produced.

The number of moles of silver the reaction produced is

15.9g⋅1 mole Ag/107.87g=0.1474 moles Ag.

This means that you'll get

0.1474moles Ag/1 mole O2/4 moles Ag=0.03685 moles  O2

Now for the tricky part. Your oxygen gas was collected over water at a total pressure of 757 mmHg. This pressure includes the pressure of the water vapor at that temperature (25∘C).

The vapor pressure of water at 25∘C is 26.69 mmHg), which means that the pressure of the oxygen will be

Ptotal=Pwater+PO2⇒PO2=Ptotal−Pwater

PO2=757−26.69=730.3 mm Hg

Now you've got all the values you need to apply the ideal gas law equation

PV=nRT⇒V=nRT P

V=0.03685moles⋅0.082atm⋅L'Mol⋅K⋅(273.15+25)K 730.3760atm

V=0.9375 L

Rounded to two sig figs, the answer will be

VO2=0.94 L

Step 2 of 3

Step 3 of 3

##### ISBN: 9780321910295

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The decomposition of a silver oxide sample forms 15.8 g of Ag(s):2 Ag2O(s) ? 4Ag(s) +