b. In an experiment, 0.06 L of 0.280 M barium hydroxide was mixed with 0.025 L of 0.300 M aluminum sulfate. Calculate the total mass of precipitates form and the molar concentrations of the salt that remain in the solution after the reaction is complete.

Alright, I need to solve this chemistry problem step by step. Let's break it down: We have two solutions being mixed: 1. 0.06 liters of 0.280 M barium hydroxide (Ba(OH)₂) 2. 0.025 liters of 0.300 M aluminum sulfate (Al₂(SO₄)₃) First, I need to determine the number of moles of each reactant. For barium hydroxide: Molarity (M) = moles/L So, moles = Molarity × Volume (in liters) Moles of Ba(OH)₂ = 0.280 M × 0.06 L = 0.0168 moles For aluminum sulfate: Moles of Al₂(SO₄)₃ = 0.300 M × 0.025 L = 0.0075 moles Now, I need to write the balanced chemical equation for the reaction between barium hydroxide and aluminum sulfate. The reaction is: Ba(OH)₂ + Al₂(SO₄)₃ → 2BaSO₄ + 2Al(OH)₃ From the balanced equation, the molar ratio is: 1 mole Ba(OH)₂ reacts with 1 mole Al₂(SO₄)₃ to produce 2 moles BaSO₄ and 2 moles Al(OH)₃ Next, I'll determine the limiting reactant. To do this, I'll compare the mole ratio of the reactants to the stoichiometric coefficients in the balanced equation. Moles of Ba(OH)₂ available = 0.0168 moles Moles of Al₂(SO₄)₃ available = 0.0075 moles According to the balanced equation: 1 mole Ba(OH)₂ reacts with 1 mole Al₂(SO₄)₃ So, 0.0075 moles of Al₂(SO₄)₃ would require 0.0075 moles of Ba(OH)₂ to react completely. Since we have 0.0168 moles of Ba(OH)₂, which is more than enough to react with 0.0075 moles of Al₂(SO₄)₃, aluminum sulfate is the limiting reactant. Now, I'll calculate the moles of precipitates formed. From the balanced equation: 1 mole Al₂(SO₄)₃ produces 2 moles BaSO₄ and 2 moles Al(OH)₃ So, 0.0075 moles Al₂(SO₄)₃ will produce: Moles of BaSO₄ = 2 × 0.0075 = 0.015 moles Moles of Al(OH)₃ = 2 × 0.0075 = 0.015 moles Total moles of precipitates = 0.015 moles BaSO₄ + 0.015 moles Al(OH)₃ = 0.030 moles Next, I'll calculate the total mass of the precipitates formed. First, find the molar masses: - BaSO₄: 137.33 (Ba) + 32.07 (S) + 4×16.00 (O) = 233.40 g/mol - Al(OH)₃: 26.98 (Al) + 3×16.00 (O) + 3×1.008 (H) = 78.00 g/mol Mass of BaSO₄ = 0.015 moles × 233.40 g/mol = 3.501 g Mass of Al(OH)₃ = 0.015 moles × 78.00 g/mol = 1.170 g Total mass of precipitates = 3.501 g + 1.170 g = 4.671 g Now, I'll determine the molar concentrations of the remaining salt in the solution after the reaction is complete. Since aluminum sulfate is the limiting reactant, all of it reacts, and barium hydroxide is in excess. Moles of Ba(OH)₂ remaining = 0.0168 moles - 0.0075 moles = 0.0093 moles Total volume of the solution after mixing = 0.06 L + 0.025 L = 0.085 L Molarity of remaining Ba(OH)₂ = moles / volume = 0.0093 moles / 0.085 L ≈ 0.110 M So, the molar concentration of the remaining salt (Ba(OH)₂) in the solution is approximately 0.110 M.