Final answer:
The amount of solid AgCl produced is equal to the amount of AgNO₃ and NaCl that react, which is 429.96 mg in this case.
Step-by-step explanation:
When 30.0 mL of 0.10 M AgNO₃ is added to 30.0 mL of 0.10 M NaCl, a double-replacement reaction occurs. The balanced equation for the reaction is: AgNO₃ (aq) + NaCl (aq) → AgCl (s) + NaNO₃ (aq). According to the stoichiometry of the reaction, 1 mole of AgNO₃ reacts with 1 mole of NaCl to produce 1 mole of AgCl. Therefore, the amount of solid AgCl produced in this reaction is equal to the amount of AgNO₃ and NaCl that react.
Concentration of AgNO₃:
30.0 mL of 0.10 M AgNO₃ contains 30.0 * 0.10 = 3.00 mmol of AgNO₃
Mass of AgCl:
1 mole of AgCl has a molar mass of 143.32 g/mol. So, the mass of AgCl is 3.00 mmol * 143.32 g/mol = 429.96 mg.
When 30.0 mL of 0.10 M AgNO₃ is mixed with 30.0 mL of 0.10 M NaCl, they undergo a double-replacement reaction to form a precipitate of AgCl (silver chloride) and aqueous NaNO₃ (sodium nitrate). The balanced chemical equation for this reaction is:
AgNO₃ (aq) + NaCl (aq) → AgCl (s) + NaNO₃ (aq)
Since the reaction is a 1:1 stoichiometry, the moles of AgCl formed will be equal to the limiting reactant's moles. Since both reactants have the same molarity and volume, neither is limiting, and all reactants will be completely consumed. To find the amount of AgCl formed, we need to calculate the moles of AgCl and then convert to mass