1. Write the balanced chemical equation:

The reaction between silver nitrate (AgNO₃) and chloride ions (Cl⁻) produces silver chloride (AgCl) as a precipitate:

AgNO₃(aq) + Cl⁻(aq) → AgCl(s) + NO₃⁻(aq)

2. Calculate the moles of silver nitrate:

Molarity (M) = moles (mol) / volume (L)

moles (mol) = Molarity (M) * volume (L)

moles of AgNO₃ = 0.141 M * 1.85 L = 0.26065 mol

3. Determine the limiting reactant:

Since the problem provides the amount of silver nitrate, we assume it's the limiting reactant. This means that the amount of silver chloride produced will be determined by the amount of silver nitrate present.

4. Use the mole ratio to find moles of silver chloride:

From the balanced equation, 1 mole of AgNO₃ produces 1 mole of AgCl. Therefore:

moles of AgCl = moles of AgNO₃ = 0.26065 mol

5. Calculate the mass of silver chloride:

mass (g) = moles (mol) * molar mass (g/mol)

molar mass of AgCl = 107.87 g/mol (Ag) + 35.45 g/mol (Cl) = 143.32 g/mol

mass of AgCl = 0.26065 mol * 143.32 g/mol = 37.3 g

Therefore, 37.3 grams of silver chloride can be produced from 1.85 L of a 0.141 M silver nitrate solution.