Question

On heating \(2.76\,\text{g}\) of \( \text{Ag}_2\text{CO}_3 (s) \), some solid residue is left behind. Determine the mass of residue left.

• A. \(2.16\,\text{g}\)
• B. \(4.32\,\text{g}\)
• C. \(2.32\,\text{g}\)
• D. \(1.08\,\text{g}\)

Answer 1

The Correct Option is A

Concept:
When silver carbonate is heated, it decomposes to form solid silver along with gaseous carbon dioxide and oxygen. The residue left after heating will therefore be metallic silver. Using stoichiometry and molar mass relations, we can determine the amount of residue formed. \[ \text{Ag}_2\text{CO}_3 (s) \xrightarrow{\Delta} 2\text{Ag}(s) + \text{CO}_2(g) + \tfrac{1}{2}\text{O}_2(g) \] Step 1: Calculate moles of \( \text{Ag}_2\text{CO}_3 \). Molar mass of \( \text{Ag}_2\text{CO}_3 \): \[ 2(108) + 12 + 3(16) = 276\,\text{g mol}^{-1} \] \[ \text{Moles of } \text{Ag}_2\text{CO}_3 = \frac{2.76}{276} = \frac{1}{100}\,\text{mol} \] Step 2: Use stoichiometry of the reaction. From the balanced equation: \[ 1\,\text{mol } \text{Ag}_2\text{CO}_3 \rightarrow 2\,\text{mol Ag} \] Therefore, \[ \frac{1}{100}\,\text{mol } \text{Ag}_2\text{CO}_3 \rightarrow \frac{2}{100}\,\text{mol Ag} \] Step 3: Calculate mass of residue (Ag). Molar mass of Ag \(=108\,\text{g mol}^{-1}\) \[ \text{Mass of Ag} = \frac{2}{100}\times108 = 2.16\,\text{g} \] Thus, the mass of solid residue left after heating is: \[ \boxed{2.16\,\text{g}} \]