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

Question

Quizplus · Accepted Answer

First, calculate the moles of AgNO3 using its concentration and volume: $0.315 \, \text{mol/L} \times 0.04000 \, \text{L} = 0.0126 \, \text{mol}$. Since the reaction requires 2 moles of AgNO3 for every mole of Na2S, the moles of Na2S needed are $0.0126 \, \text{mol} / 2 = 0.0063 \, \text{mol}$. To find the volume of Na2S solution required, use its concentration: $0.0063 \, \text{mol} / 0.260 \, \text{mol/L} = 0.0242 \, \text{L}$, or 24.2 mL.

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