Question

B. Solid silver oxide decomposes at temperatures in excess of 300°C, yielding metallic silver and oxygen gas. A 3.13 g sample of impure silver oxidel yields 0.187 g oxygen. If silver oxide is the only source of O2, what is the percent silver oxide by mass in the sample? (15 pts.)

Answer 1

Solution:

Given, Mass of impure Ag2O = 3.13 gm

Mass of O2 formed = 0.187 gm

We need to calculate percent Ag2O by mass in the sample.

The reaction of Ag2O is:

2 Ag_{​​​​​2}​​​​O -----> 4 Ag + O2

We can see that 2 moles of Ag2O formed 1 mole of O2.

We know, 32 gm of O2 = 22.4 L

Or, 1 gm of O2 = ( 22.4 / 32 )L

So, 0.187 gm of O2 = ( 22.4 / 32 ) × ( 0.187 ) L

Or, Volume of O2 formed = 0.13 L

We know, Molar mass of Ag2O = 232 gm

As 2 moles of Ag2O have reacted, so we can say that

22.4 L of O2 = ( 2 × 232 ) gm Ag2O reacted

Or, 1 L of O2 = ( 464 gm / 22.4 L )

So,. 0.13 L of O2 = ( 464 gm / 22.4 L ) × ( 0.13 L )

Or, Actual mass of Ag2O reacted = 2.7 gm

We know,

Percent by mass = ( Actual mass / Impure mass ) × 100

So, Percent Ag2O by mass = ( 2.7 gm / 3.13 gm ) × 100

Or, Percent Ag2O by mass = 86.26 %

Hence, Percent Ag2O by mass in sample = 86.26%