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Chemistry Question 51 – JEE-MAIN 2026

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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)

Identify the correct balanced chemical equation for the thermal decomposition of silver carbonate.

Step 1: Write the balanced decomposition reaction and calculate molar mass✦ Active

Silver carbonate (Ag2CO3) decomposes upon heating to form silver metal (Ag), carbon dioxide (CO2), and oxygen gas (O2). The residue obtained is silver metal.

2Ag2CO3(s)→Δ4Ag(s)+2CO2(g)+O2(g)

Given molar masses: Ag=108 g mol−1, C=12 g mol−1, O=16 g mol−1. Molar mass of Ag2CO3=(2×108)+12+(3×16)=216+12+48=276 g mol−1.

Step 2: Calculate moles of silver carbonate and silver produced○ Expand

Given mass of silver carbonate = 2.76 g. Moles of Ag2CO3=MassMolar mass=2.76 g276 g mol−1=0.01 mol.

From the balanced chemical equation, 2 moles of Ag2CO3 produce 4 moles of Ag. Therefore, 0.01 mol of Ag2CO3 will produce: 0.01 mol Ag2CO3×4 mol Ag2 mol Ag2CO3=0.02 mol Ag.

Step 3: Calculate the mass of silver (residue)○ Expand

The molar mass of silver (Ag) is 108 g mol−1. Mass of Ag=Moles of Ag×Molar mass of Ag=0.02 mol×108 g mol−1=2.16 g.

💡 Teacher's Secret Hint

Ensure to use the correct molar mass for silver metal, which is the residue.

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