Thermal decomposition of Group 2 nitrate to oxide residue Steps:

• Reaction: M(NO3)2 → MO + 2NO2 + ½O2; residue is MO.
• Molar mass of M(NO3)2 = M + 124; molar mass of MO = M + 16.
• Ratio of residue mass to initial mass = (M + 16)/(M + 124) = 3.7/4 = 0.925.
• Solve 0.925(M + 124) = M + 16 → M ≈ 1316 (does not match any Group 2 metal).

Why B is correct:

• Not enough information (calculated M does not match calcium or any option).

Why the others are wrong:

• A. Barium (M=137) gives ratio 153/261 ≈ 0.586; expected residue 2.3 g.
• C. Magnesium (M=24) gives ratio 40/148 ≈ 0.270; expected residue 1.1 g.
• D. Strontium (M=88) gives ratio 104/212 ≈ 0.491; expected residue 2.0 g.

Final answer: Not enough information.