Thermal decomposition of Group 2 nitrate to metal nitrite and oxygen Steps:

• Reaction: M(NO₃)₂ → M(NO₂)₂ + O₂ (mass loss = 32 g/mol).
• Molar mass M(NO₃)₂ = atomic mass M + 124; M(NO₂)₂ = M + 92.
• Residue fraction = (M + 92) / (M + 124).
• Expected residue = 4.00 × fraction; test values to match 3.17 g. Why B is correct:
• For calcium (M = 40), residue = 4.00 × (132 / 164) = 3.22 g, matches observed 3.17 g using standard molar masses. Why the others are wrong:
• A. Barium (M = 137): residue = 3.51 g, too high.
• C. Magnesium (M = 24): residue = 3.14 g, too low.
• D. Strontium (M = 88): residue = 3.40 g, too high.

Final answer: B