Answer to Question #316503 in General Chemistry for Autumn123

Question

If 0.650% solution of Na2CO3 will be used to precipitate Ca+² from solution, determine the amount of solution,in grams, needed to precipitate 5.00×10² mL of 0.01120M Ca+²?

Answer 1

5.00×10² mL = 5.00×10^-1 L

Moles of Ca²⁺: 5.00 × 10^-1 L * 0.01120 M = 0.0056 mol

Ca²⁺ + CO₃^2- = CaCO₃

The mole ratio is 1 : 1, So, 0.0056 mol of Na₂CO₃ is used for precipitation

Molar mass Na₂CO₃: 106 g/mol

Mass Na₂CO₃: 0.0056 * 106 = 0.5936 g

Mass of solution:

0.5936 / 0.0065 = 91.3 g