1.12 g of solid potassium hydroxide (KOH) is added to 50 mL of a 0.25 M aqueous solution of H ₂ SO ₄ . Assuming complete neutralization, the mass of the unreacted reactant left in the mixture is: (Given molar masses in g mol ⁻¹ : K = 39, O = 16, H = 1, S = 32)

Step-by-step solution

Molar mass of KOH = 39 + 16 + 1 = 56 g mol ⁻¹ Moles of KOH = 1.12 56 = 0.02 mol = 20 mmol Moles of H ₂ SO ₄ = 50 0.25 = 12.5 mmol The balanced chemical equation is: 2 KOH + H ₂ SO ₄ K ₂ SO ₄ + 2 H ₂ O From the stoichiometry, 2 moles of KOH react with 1 mole of H ₂ SO ₄ . Moles of H ₂ SO ₄ required to react with 20 mmol of KOH = 20 2 = 10 mmol Since 12.5 mmol of H ₂ SO ₄ is present, KOH is the limiting reagent and H ₂ SO ₄ is in excess. Unreacted H ₂ SO ₄ = 12.5 - 10 = 2.5 mmol Molar mass of H ₂ SO ₄ = 2(1) + 32 + 4