Percentage yield from limiting reactant

Steps:

• Moles of Na₂SO₄ = (25/1000) dm³ × 1.0 mol dm⁻³ = 0.025 mol.
• Theoretical moles of BaSO₄ = 0.025 mol (1:1 stoichiometry with excess BaCl₂).
• Theoretical mass of BaSO₄ = 0.025 mol × 233 g mol⁻¹ = 5.825 g.
• Percentage yield = (5.36 g / 5.825 g) × 100% = 92%.

Why C is correct:

• Percentage yield formula (actual mass / theoretical mass) × 100% yields exactly 92% based on molar mass of BaSO₄ (233 g mol⁻¹).

Why the others are wrong:

• A: Error in volume conversion, treating 25 cm³ as 25 dm³.
• B: Divides actual mass by molar mass without moles calculation.
• D: Forgets to convert cm³ to dm³, yielding ~9% from incorrect moles.

Final answer: C