When 10.0 g of AgNO3 is added to an aqueous solution containing excess NaCl, what mass of AgCl will precipitate? (Assume complete reaction and that AgNO3 is the limiting reagent.)

• A. 3.94 g
• B. 5.87 g
• C. 8.44 g
• D. 10.0 g

Answer: (C)

Verified step by step guidance

1. Write the balanced chemical equation for the reaction between silver nitrate (AgNO\_3) and sodium chloride (NaCl): \[\text{AgNO\_3 (aq)} + \text{NaCl (aq)} \rightarrow \text{AgCl (s)} + \text{NaNO\_3 (aq)}\]
2. Calculate the number of moles of AgNO\_3 used. Use the molar mass of AgNO\_3, which can be found by summing the atomic masses of Ag, N, and O\_3: \[\text{moles of AgNO\_3} = \frac{\text{mass of AgNO\_3}}{\text{molar mass of AgNO\_3}}\]
3. Since AgNO\_3 is the limiting reagent and the reaction is 1:1, the moles of AgCl formed will be equal to the moles of AgNO\_3 reacted: \[\text{moles of AgCl} = \text{moles of AgNO\_3}\]
4. Calculate the mass of AgCl precipitated by multiplying the moles of AgCl by its molar mass (sum of atomic masses of Ag and Cl): \[\text{mass of AgCl} = \text{moles of AgCl} \times \text{molar mass of AgCl}\]
5. This final mass is the amount of AgCl that will precipitate when 10.0 g of AgNO\_3 reacts completely with excess NaCl.