Question

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?

Short answer

Answer: The expected precipitate is barium sulfate (BaSO4), and the mass of this precipitate is 10.75 g. The chloride concentration after precipitation is complete is 1.23 M.

Step-by-step solution

(Step 1: Write the balanced chemical equation)

(The balanced chemical equation for the reaction between sulfuric acid (H2SO4) and barium chloride (BaCl2) is: H2SO4 (aq) + BaCl2 (aq) -> BaSO4 (s) + 2HCl (aq) The expected precipitate is barium sulfate (BaSO4).)

(Step 2: Calculate the moles of H2SO4 and BaCl2)

(First, we need to calculate the moles of H2SO4 and BaCl2. To find the number of moles of H2SO4, first convert the mass percentage (15.25%) to mass, then divide by the molar mass of H2SO4. Moles of H2SO4 = (Volume * Density of solution * mass percentage of H2SO4 / Molar mass of H2SO4) Moles of H2SO4 = (25 mL * 1.107 g/mL * 0.1525) / 98.08 g/mol = 0.04609 mol To find the number of moles of BaCl2, use the formula: Moles of BaCl2 = Molarity * Volume Moles of BaCl2 = 2.45 M * 0.050 L = 0.1225 mol)

(Step 3: Determine the limiting reactant)

(Using the stoichiometric coefficients from the balanced equation, we can determine which reactant is the limiting reactant, which is the one that will finish first during the reaction: Moles of H2SO4 = 0.04609 mol Moles of BaCl2 = 0.1225 mol Since 1 mole of H2SO4 reacts with 1 mole of BaCl2, then the limiting reactant is H2SO4, as there are fewer moles available for the reaction.)

(Step 4: Calculate the moles of precipitate formed)

(Now, using the stoichiometry of the reaction, we can determine the moles of BaSO4 formed: Moles of BaSO4 = Moles of limiting reactant = Moles of H2SO4 = 0.04609 mol)

(Step 5: Calculate the grams of precipitate formed)

(To find the mass of precipitate formed, multiply the moles of BaSO4 by the molar mass of BaSO4: Mass of BaSO4 = Moles of BaSO4 * Molar mass of BaSO4 Mass of BaSO4 = 0.04609 mol * 233.43 g/mol = 10.75 g)

(Step 6: Calculate the moles of HCl formed)

(As per the balanced chemical equation, 2 moles of HCl are formed for every 1 mole of H2SO4 reacted: Moles of HCl = 2 * Moles of H2SO4 = 2 * 0.04609 = 0.09218 mol)

(Step 7: Calculate the concentration of chloride ions after precipitation is complete)

(Now that we have the moles of HCl, we can calculate the concentration of chloride ions after precipitation is complete. Since the initial volume of BaCl2 solution is 50 mL and the initial volume of H2SO4 solution is 25 mL, the final volume is 75 mL (0.075 L). The concentration of chloride ions is given by: [Cl⁻] = (Moles of HCl) / (Final volume) [Cl⁻] = 0.09218 mol / 0.075 L = 1.23 M) So to summarize: a) The expected precipitate is barium sulfate (BaSO4). b) The grams of precipitate obtained is 10.75 g. c) The chloride concentration after precipitation is complete is 1.23 M.