1. Write the balanced chemical equation:

AgNO₃ (aq) + BaCl₂ (aq) → 2AgCl (s) + Ba(NO₃)₂ (aq)

2. Determine the limiting reactant:

* Calculate the moles of each reactant:

* Moles of AgNO₃ = (10.0 g) / (169.87 g/mol) = 0.0589 mol

* Moles of BaCl₂ = (15.0 g) / (208.23 g/mol) = 0.0720 mol

* Compare the mole ratio: The balanced equation shows a 1:1 mole ratio between AgNO₃ and BaCl₂. Since we have fewer moles of AgNO₃, it is the limiting reactant.

3. Calculate the moles of AgCl formed:

* From the balanced equation, 1 mole of AgNO₃ produces 2 moles of AgCl.

* Moles of AgCl = (0.0589 mol AgNO₃) * (2 mol AgCl / 1 mol AgNO₃) = 0.118 mol AgCl

4. Calculate the mass of AgCl formed:

* Mass of AgCl = (0.118 mol) * (143.32 g/mol) = 16.9 g AgCl

Therefore, 16.9 grams of silver chloride (AgCl) are formed.