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:
The topic is Stoichiometry, involving a thermal decomposition reaction.
We need to calculate the mass of the solid product (residue) formed when a given mass of silver carbonate (\(\text{Ag}_2\text{CO}_3\)) is heated.
Step 2: Key Formula or Approach:
The first step is to write the correct balanced chemical equation for the decomposition. Silver carbonate decomposes to metallic silver, carbon dioxide, and oxygen. Then, we use the mole concept to relate the mass of reactant to the mass of the product.
Balanced Equation: \[ 2\text{Ag}_2\text{CO}_3(s) \xrightarrow{\Delta} 4\text{Ag}(s) + 2\text{CO}_2(g) + \text{O}_2(g) \]
Or simplified: \[ \text{Ag}_2\text{CO}_3 \xrightarrow{\Delta} 2\text{Ag} + \text{CO}_2 + \frac{1}{2}\text{O}_2 \]
Step 3: Detailed Explanation:
1. Calculate Molar Mass of \(\text{Ag}_2\text{CO}_3\):
Molar mass = \(2 \times (\text{Ag}) + 1 \times (\text{C}) + 3 \times (\text{O})\)
= \((2 \times 108) + 12 + (3 \times 16) = 216 + 12 + 48 = 276 \text{ g/mol}\).
2. Calculate Moles of \(\text{Ag}_2\text{CO}_3\):
Moles = \(\frac{\text{Given Mass}}{\text{Molar Mass}} = \frac{2.76 \text{ g}}{276 \text{ g/mol}} = 0.01 \text{ mol}\).
3. Use Stoichiometry to find Moles of Residue (\(\text{Ag}\)):
The solid residue is metallic silver (\(\text{Ag}\)). From the balanced equation, 1 mole of \(\text{Ag}_2\text{CO}_3\) produces 2 moles of \(\text{Ag}\).
Therefore, \(0.01 \text{ mol}\) of \(\text{Ag}_2\text{CO}_3\) will produce \(2 \times 0.01 = 0.02 \text{ mol}\) of \(\text{Ag}\).
4. Calculate Mass of Ag Residue:
Mass = Moles \(\times\) Molar Mass of Ag
Mass = \(0.02 \text{ mol} \times 108 \text{ g/mol} = 2.16 \text{ g}\).
Step 4: Final Answer:
The mass of residue obtained is \(2.16 \text{ g}\).