When 200 mL of 0.10 M BaCl₂(aq)is added to 100 mL of 0.30 M Na₂SO₄(aq),the number of moles of BaSO₄(solubility product = 1.1 × 10^-10)precipitated is ________. A)0.020 mol B)0.010 mol C)0.20 mol D)0.030 mol E)0.10 mol

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The reaction between BaCl2 and Na2SO4 forms BaSO4 as a precipitate. Initially, we have 0.020 mol of Ba2+ (from 200 mL of 0.10 M BaCl2) and 0.030 mol of SO4^2- (from 100 mL of 0.30 M Na2SO4). Since BaSO4 has a 1:1 stoichiometry and is limited by the lesser amount of reactant, which is Ba2+, all 0.020 mol of Ba2+ will react to form BaSO4 precipitate.

When 200 ML of 0