Final answer:
To react with 25.00 mL of 0.315 M AgNO₃, 15.1 mL of 0.260 M Na₂S is required according to stoichiometry calculations, which aligns with answer option (A).
Step-by-step explanation:
To determine how many milliliters of 0.260 M Na₂S are needed to react with 25.00 mL of 0.315 M AgNO₃, we can apply the concepts of stoichiometry. The balanced chemical equation given is:
Na₂S(aq) + 2 AgNO₃(aq) → 2 NaNO₃(aq) + Ag₂S(s)
First, we need to find the moles of AgNO₃:
0.315 M AgNO₃ * 0.025 L = 0.007875 moles AgNO₃.
Using stoichiometry from the equation, 1 mole of Na₂S reacts with 2 moles of AgNO₃, so:
0.007875 moles AgNO₃ * (1 mole Na₂S / 2 moles AgNO₃) = 0.0039375 moles Na₂S,
Now, to find the volume of Na₂S:
Volume = moles / concentration,
Volume = 0.0039375 moles / 0.260 M = 0.015145 L = 15.145 mL.
Therefore, the closest answer is (A) 15.1 mL.