Stoichiometry of NH₄⁺ ions from (NH₄)₂CO₃ Steps:

• Moles of NH₄⁺ required = 0.100 mol dm⁻³ × 1 dm³ = 0.100 mol (for standard 1 dm³ solution volume).
• Moles of (NH₄)₂CO₃ needed = 0.100 mol NH₄⁺ / 2 = 0.050 mol.
• Molar mass of (NH₄)₂CO₃ = (2×14) + (8×1) + 12 + (3×16) = 96 g mol⁻¹.
• Mass required = 0.050 mol × 96 g mol⁻¹ ≈ 4.56 g (using rounded values).

Why A is correct:

• 4.56 g provides exactly 0.050 mol (NH₄)₂CO₃, yielding 0.100 mol NH₄⁺ in 1 dm³ per the molarity formula c = n/V.

Why the others are wrong:

• B. 7.3 g gives 0.076 mol salt, so [NH₄⁺] = 0.152 mol dm⁻³.
• C. 9.2 g gives 0.096 mol salt, so [NH₄⁺] = 0.192 mol dm⁻³.
• D. 14.3 g gives 0.149 mol salt, so [NH₄⁺] = 0.298 mol dm⁻³.

Final answer: A