JEE Main 2026 — Apr 5, Shift 1 — Chemistry solution

JEE Main2026Apr 5, Shift 1Chemistry

How many grams of residue is obtained by heating 2.76 g of silver carbonate ?

(Given : Molar mass of C, O and Ag are 12, 16 and 108 g mol $^{- 1}$ respectively)

Solution

On heating, silver carbonate decomposes to metallic silver, carbon dioxide, and oxygen:

$Ag_{2} CO_{3} (s) Δ 2 Ag (s) + CO_{2} (g) + \frac{1}{2} O_{2} (g)$ .

The non-volatile residue is silver metal. Molar mass of $Ag_{2} CO_{3} = 2 (108) + 12 + 3 (16) = 276$ g/mol.

Moles of $Ag_{2} CO_{3} = \frac{2.76}{276} = 0.01$ mol, giving $0.02$ mol of Ag.

Mass of residue $= 0.02 \times 108 = 2.16$ g.

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