What mass of an 80 % pure BaCl ₂ sample is required to completely precipitate the sulfate ions from 50 mL of a 0.2 M Na ₂ SO ₄ solution? (Given molar mass of BaCl ₂ = 208 g mol ⁻¹ )

Step-by-step solution

The balanced chemical equation for the precipitation reaction is: BaCl ₂( aq ) + Na ₂ SO ₄( aq ) BaSO ₄( s ) + 2 NaCl ( aq ) Number of moles of Na ₂ SO ₄ taken = Molarity Volume in L = 0.2 M 0.050 L = 0.01 mol From the balanced equation, 1 mole of Na ₂ SO ₄ reacts with 1 mole of BaCl ₂ . Therefore, moles of pure BaCl ₂ required = 0.01 mol . Mass of pure BaCl ₂ required = 0.01 mol 208 g mol ⁻¹ = 2.08 g . Since the sample is 80 % pure, the mass of the impure sample required is: Mass of impure sample = Mass of pure su