1. Write the balanced chemical equation:

The reaction between silver nitrate (AgNO₃) and calcium chloride (CaCl₂) produces silver chloride (AgCl) precipitate and calcium nitrate (Ca(NO₃)₂):

2 AgNO₃ (aq) + CaCl₂ (aq) → 2 AgCl (s) + Ca(NO₃)₂ (aq)

2. Calculate the moles of each reactant:

* AgNO₃:

* Molar mass of AgNO₃ = 169.87 g/mol

* Moles of AgNO₃ = 14 g / 169.87 g/mol = 0.0824 mol

* CaCl₂:

* Molar mass of CaCl₂ = 110.98 g/mol

* Moles of CaCl₂ = 4.83 g / 110.98 g/mol = 0.0435 mol

3. Determine the limiting reactant:

The limiting reactant is the one that gets consumed completely first, limiting the amount of product formed.

* From the balanced equation, 2 moles of AgNO₃ react with 1 mole of CaCl₂.

* The mole ratio of AgNO₃ to CaCl₂ is 2:1.

* We have 0.0824 mol of AgNO₃, which would require 0.0824 mol / 2 = 0.0412 mol of CaCl₂.

* Since we have more CaCl₂ (0.0435 mol) than needed, AgNO₃ is the limiting reactant.

4. Calculate the moles of precipitate (AgCl):

* From the balanced equation, 2 moles of AgNO₃ produce 2 moles of AgCl.

* The mole ratio of AgNO₃ to AgCl is 1:1.

* Therefore, 0.0824 mol of AgNO₃ will produce 0.0824 mol of AgCl.

5. Calculate the mass of precipitate (AgCl):

* Molar mass of AgCl = 143.32 g/mol

* Mass of AgCl = 0.0824 mol * 143.32 g/mol = 11.78 g

Therefore, the mass of the precipitate (AgCl) produced is approximately 11.78 grams.