Twenty-five milliliters of a solution \((d=1.107 \mathrm{~g} / \mathrm{mL})\) containing \(15.25 \%\) by mass of sulfuric acid is added to \(50.0 \mathrm{~mL}\) of \(2.45 \mathrm{M}\) barium chloride. (a) What is the expected precipitate? (b) How many grams of precipitate are obtained? (c) What is the chloride concentration after precipitation is complete?

In this mixture, we have two solutions: a solution containing sulfuric acid (H2SO4) and a solution containing barium chloride (BaCl2). When the two solutions are mixed, the following double displacement reaction takes place: H2SO4 + BaCl2 -> 2HCl + BaSO4 Barium sulfate (BaSO4) is insoluble in water and will precipitate during the reaction.

First, let's determine the amount of moles of H2SO4 in the sulfuric acid solution: Density of the solution is given as 1.107 g/mL and the volume is 25 mL. So the mass of the solution would be: mass_solution = (1.107 g/mL) x 25 mL = 27.675 g The solution contains 15.25% by mass of H2SO4, mass_H2SO4 = 0.1525 x 27.675 g = 4.222 g The molar mass of H2SO4 is 98.08 g/mol, so the moles of H2SO4 is given by: moles_H2SO4 = 4.222 g / 98.08 g/mol = 0.0430 mol Now let's determine the amount of moles of BaCl2 in the barium chloride solution: The concentration of the solution is 2.45 M, and the volume is 50.0 mL, so the moles of BaCl2 are: moles_BaCl2 = 2.45 M x 50.0 mL x (1 L/1000 mL) = 0.1225 mol From the balanced chemical equation, the mole ratio of H2SO4 to BaCl2 is 1:1. Since we have less moles of H2SO4 (0.0430 mol) than BaCl2 (0.1225 mol), H2SO4 is the limiting reactant.

The mole ratio of H2SO4 to BaSO4 is 1:1. So the moles of BaSO4 formed will be equal to the moles of the limiting reactant, H2SO4. moles_BaSO4 = 0.0430 mol The molar mass of BaSO4 is 233.43 g/mol, so the mass of the BaSO4 precipitate is given by: mass_BaSO4 = moles_BaSO4 x molar_mass_BaSO4 = (0.0430 mol)(233.43 g/mol) = 10.04 g

The mole ratio of BaCl2 to BaSO4 is 1:1. So the remaining moles of BaCl2 after the reaction is: remaining_moles_BaCl2 = initial_moles_BaCl2 - moles_BaSO4 = 0.1225 mol - 0.0430 mol = 0.0795 mol Since the volume of the final mixture is 25 mL of the sulfuric acid solution and 50.0 mL of the barium chloride solution, the total volume is 75.0 mL. Therefore, the remaining chloride concentration is: Cl_concentration = remaining_moles_BaCl2 / final_volume = 0.0795 mol / 75.0 mL x (1000 mL / 1 L) = 1.06 M Summary: (a) The expected precipitate is barium sulfate (BaSO4). (b) The amount of precipitate obtained is 10.04 g. (c) The chloride concentration after precipitation is 1.06 M.