1. Write the balanced chemical equation:

NaCl (aq) + AgNO₃ (aq) → AgCl (s) + NaNO₃ (aq)

2. Determine the limiting reactant:

Since we have excess AgNO₃, NaCl is the limiting reactant. This means the amount of AgCl produced is determined by the amount of NaCl available.

3. Calculate the moles of NaCl:

* Molar mass of NaCl = 58.44 g/mol

* Moles of NaCl = (10 g) / (58.44 g/mol) = 0.171 mol

4. Use the mole ratio from the balanced equation to find moles of AgCl:

* The balanced equation shows a 1:1 mole ratio between NaCl and AgCl.

* Therefore, 0.171 mol of NaCl will produce 0.171 mol of AgCl.

5. Calculate the mass of AgCl:

* Molar mass of AgCl = 143.32 g/mol

* Mass of AgCl = (0.171 mol) * (143.32 g/mol) = 24.5 g

Therefore, 24.5 grams of solid AgCl are produced in this reaction.