When 10.0 g of NaCl is added to an aqueous solution containing excess AgNO3, what mass of AgCl will precipitate?

• A. 8.45 g
• B. 10.0 g
• C. 5.85 g
• D. 14.7 g

Answer: (D)

Verified step by step guidance

1. Write the balanced chemical equation for the reaction between sodium chloride (NaCl) and silver nitrate (AgNO\_3): \(\mathrm{NaCl (aq) + AgNO_3 (aq) \rightarrow AgCl (s) + NaNO_3 (aq)}\).
2. Calculate the number of moles of NaCl used by dividing its mass by its molar mass: \(\text{moles of NaCl} = \frac{10.0\ \mathrm{g}}{\text{molar mass of NaCl}}\) where the molar mass of NaCl is approximately 58.44 g/mol.
3. Use the stoichiometry of the reaction to find the moles of AgCl formed. Since the reaction is a 1:1 molar ratio, moles of AgCl formed = moles of NaCl reacted.
4. Calculate the mass of AgCl precipitated by multiplying the moles of AgCl by its molar mass: \(\text{mass of AgCl} = \text{moles of AgCl} \times \text{molar mass of AgCl}\), where the molar mass of AgCl is approximately 143.32 g/mol.
5. Interpret the result as the mass of AgCl that will precipitate when 10.0 g of NaCl reacts with excess AgNO\_3.