Answer to Question #325156 in General Chemistry for Emmer

Question

Determine the minimum volume of 0.50 mol/L calcium chloride solution required to precipitate 14 g of calcium carbonate when mixed with excess sodium carbonate

Answer 1

CaCl₂ + Na₂CO₃ = CaCO₃ + 2NaCl

n(CaCO₃) = m(CaCO₃) / Mr(CaCO₃) = 14 g / 100 g/mol = 0.14 mol

1 mol CaCl₂ - 1 mol CaCO₃

x mol CaCl₂ - 0.14 mol CaCO₃

x = 0.14 mol

V(sol) = n(CaCO₃) / C(sol) = 0.14 mol / 0.50 mol/l = 0.28 l = 280 ml