Question

How many grams of residue is obtained by heating \(2.76 \text{ g}\) of silver carbonate? (Given : Molar mass of C, O and Ag are 12, 16 and \(108 \text{ g mol}^{-1}\) respectively)

• A. \(1.08 \text{ g}\)
• B. \(2.16 \text{ g}\)
• C. \(3.24 \text{ g}\)
• D. \(4.32 \text{ g}\)

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
Silver carbonate (\(\text{Ag}_2\text{CO}_3\)) decomposes upon heating. Unlike most other carbonates that form metal oxides, silver carbonate eventually decomposes to metallic silver because silver oxide is itself thermally unstable at high temperatures.
Step 2: Key Formula or Approach:
The decomposition reaction is:
\[ \text{Ag}_2\text{CO}_3 \xrightarrow{\Delta} 2\text{Ag} + \text{CO}_2 + \frac{1}{2}\text{O}_2 \]
The solid residue remaining is metallic silver (\(\text{Ag}\)).
Step 3: Detailed Explanation:
1. Calculate the molar mass of \(\text{Ag}_2\text{CO}_3\):
\(M(\text{Ag}_2\text{CO}_3) = (2 \times 108) + 12 + (3 \times 16) = 216 + 12 + 48 = 276 \text{ g/mol}\).
2. Calculate the moles of \(\text{Ag}_2\text{CO}_3\) heated:
\[ n(\text{Ag}_2\text{CO}_3) = \frac{2.76 \text{ g}}{276 \text{ g/mol}} = 0.01 \text{ mol} \]
3. Use stoichiometry to find moles of residue (\(\text{Ag}\)):
From the balanced equation, \(1 \text{ mole}\) of \(\text{Ag}_2\text{CO}_3\) gives \(2 \text{ moles}\) of \(\text{Ag}\).
So, \(0.01 \text{ mol}\) of \(\text{Ag}_2\text{CO}_3\) gives \(2 \times 0.01 = 0.02 \text{ mol}\) of \(\text{Ag}\).
4. Calculate the mass of the \(\text{Ag}\) residue:
\[ \text{Mass of Ag} = 0.02 \text{ mol} \times 108 \text{ g/mol} = 2.16 \text{ g} \]
Step 4: Final Answer:
The mass of residue obtained is \(2.16 \text{ g}\).