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Chapter 4: Problem 24

Solid silver oxide, \(\mathrm{Ag}_{2} \mathrm{O}(\mathrm{s}),\) decomposes at temperatures in excess of \(300^{\circ} \mathrm{C},\) yielding metallic silver and oxygen gas. A 3.13 g sample of impure silver oxide yields \(0.187 \mathrm{g} \mathrm{O}_{2}(\mathrm{g}) .\) What is the mass percent \(\mathrm{Ag}_{2} \mathrm{O}\) in the sample? Assume that \(\mathrm{Ag}_{2} \mathrm{O}(\mathrm{s})\) is the only source of \(\mathrm{O}_{2}(\mathrm{g}) .\) [Hint: Write a balanced equation for the reaction.]

Short Answer

Expert verified
The mass percent of \(\mathrm{Ag}_{2} \mathrm{O}\) in the sample is approximately 86.9%.

Step by step solution

01

Writing the Balanced Chemical Reaction

From the problem, we know that solid silver oxide, \(\mathrm{Ag}_{2} \mathrm{O}\), decomposes into silver (\(\mathrm{Ag}\)) and oxygen (\(\mathrm{O}_{2}\)). This allows us to write the following unbalanced reaction: \(\mathrm{Ag}_{2} \mathrm{O}(\mathrm{s}) \rightarrow \mathrm{Ag}(\mathrm{s}) + \mathrm{O}_{2}(\mathrm{g})\). To balance the reaction, we update it to: \(2\mathrm{Ag}_{2} \mathrm{O}(\mathrm{s}) \rightarrow 4\mathrm{Ag}(\mathrm{s}) + \mathrm{O}_{2}(\mathrm{g})\).
02

Applying Stoichiometry to Find Moles

From balanced equation, we know that 2 moles of \(\mathrm{Ag}_{2} \mathrm{O}\) will give 1 mole of \(\mathrm{O}_{2}\). But we need to convert grams to moles. The molecular mass of oxygen is 32 g/mol. Hence, 0.187 g of \(\mathrm{O}_{2}\) is \(0.187/32=0.00584\) moles. As, 2 moles of \(\mathrm{Ag}_{2} \mathrm{O}\) give 1 mole of \(\mathrm{O}_{2}\), then \(0.00584\) moles of \(\mathrm{O}_{2}\) has been given by \(2*0.00584=0.0117\) moles of \(\mathrm{Ag}_{2} \mathrm{O}\).
03

Finding the Mass of \(\mathrm{Ag}_{2} \mathrm{O}\) in the Sample

Knowing the moles of \(\mathrm{Ag}_{2} \mathrm{O}\), we can calculate the mass. The molecular weight of \(\mathrm{Ag}_{2} \mathrm{O}\) is 231.74 g/mol. So, the mass of \(\mathrm{Ag}_{2} \mathrm{O}\) in the sample is \(0.0117*231.74 = 2.72\) g.
04

Finding the Mass Percent

The mass percent of \(\mathrm{Ag}_{2} \mathrm{O}\) in the sample is calculated using the mass of \(\mathrm{Ag}_{2} \mathrm{O}\) and the total mass of the sample. The mass percent is given by \((\(\mathrm{Ag}_{2} \mathrm{O}\) Mass / Total mass) * 100\), which equals to \((2.72/3.13) * 100 = 86.9\)%.

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