Answer to Question #169629 in Chemistry for Samantha Hernandez

Question

500 mL of 0.0100 M CaCl₂ is mixed with 500 mL of 0.0100 M Na₂SO₄. How many grams of CaSO₄ will precipitate when the equilibrium is reached? The solubility product constant of CaSO₄ is K_sp = 2.50 × 10-⁵. MM(CaSO₄) = 136.14 g mol^–1.

•  A. 1.36 g
•  B. 2.04 g
•  C. No precipitate will form
•  D. 0.68 g
•  E. 5.45 g

Answer 1

CaCl₂ + Na₂SO₄ = 2NaCl + CaSO₄

С_M = n/V

n = C_M x V

n = m/M

n (CaCl₂) = 0.0100 x 0.5 = 0.005 mol

n (Na₂SO₄) = 0.0100 x 0.5 = 0.005 mol

According to the equation, n (CaCl₂) = n (Na₂SO₄) = n (CaSO₄) = 0.005 mol

M (CaSO₄) = 136.14 g/mol

m (CaSO₄) = n x M = 0.005 x 136.14 = 0.68 g

Answer: D - 0.68 g