1. write a balanced equation for the precipitation of calcium carbonate from potassium carbonate and calcium chloride. 2. using this balanced equation, determine the limiting reactant if 15 grams of calcium chloride was reacted with 15 grams of potassium carbonate. 3. using your answer for question 2, determine the mass of potassium carbonate needed to fully precipitate all the calcium from a 25 ml sample of 15% calcium chloride.

1)

2) K2CO3 is the limiting reagent

3) Mass of K2CO3 need = 4.67 g

Explanation:

1) Potassium carbonate reacts with calcium chloride to form calcium carbonate and potassium chloride. The balanced equation is:

Based on the reaction stoichiometry:

1 mole of Potassium carbonate reacts with 1 mole of calcium chloride to form 1 mole of calcium carbonate and 2 moles potassium chloride

2) Mass of CaCl2 = 15 g

Molar mass of CaCl2 = 110.98 g/mol

Mass of K2CO3 = 15 g

Molar mass of K2CO3 = 138.21 g/mol

Since the moles of K2CO3 < moles of CaCl2, potassium carbonate is the limiting reagent

3) 15% CaCl2 implies: 15 g CaCl2 in 100 ml solution

Therefore in the given 25 ml solution, the mass of CaCl2 would be:

From the reaction stoichiometry the molar ratio of K2CO3:CaCl2 = 1:1

Therefore, moles of K2CO3 that would be needed to fully precipitate 0.0338 moles of CaCl2 present in 25 ml of 15% solution would be = 0.0338 moles

The corresponding mass of K2CO3 would be:

58750.08068 is the answer i think