Percentage yield from limiting reactant Steps:

• Molar masses: NH₄Cl ≈ 50 g/mol (approx.), so 2NH₄Cl ≈ 100 g; CaCl₂ = 111 g/mol; Ca(OH)₂ = 74 g/mol.
• Moles NH₄Cl = 10 / 50 = 0.2 mol; moles Ca(OH)₂ = 14 / 74 ≈ 0.19 mol.
• Ratio 2 NH₄Cl : 1 Ca(OH)₂ means 0.2 mol NH₄Cl requires 0.1 mol Ca(OH)₂ (< 0.19 mol), so NH₄Cl limiting.
• Theoretical CaCl₂ = (0.2 / 2) × 111 = 11.1 g; yield = (10 / 11.1) × 100 ≈ 91%. Why D is correct:
• Percentage yield = (actual / theoretical) × 100% matches 91% using stoichiometric mass ratio from limiting reactant. Why the others are wrong:
• A ignores stoichiometry, underestimates yield.
• B assumes Ca(OH)₂ limiting, gives ~48%.
• C misapplies moles, yields ~67% erroneously.

Final answer: D