How many millilitres of 0.260 mol L^-1 Na₂S are needed to react with 40.00 mL of 0.315 mol L^-1 AgNO₃? Na₂S(aq) + 2AgNO₃(aq) → 2NaNO₃(aq) + Ag₂S(s)

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The balanced equation shows that 1 mole of Na2S reacts with 2 moles of AgNO3. Calculate moles of AgNO3: 0.315 mol/L * 0.040 L = 0.0126 mol. Moles of Na2S needed: 0.0126 mol / 2 = 0.0063 mol. Volume of Na2S: 0.0063 mol / 0.260 mol/L = 0.0242 L = 24.2 mL.

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