How many milliliters of 0.260 M Na₂S are needed to react with 45.00 mL of 0.315 M AgNO₃? Na₂S(aq)+ 2 AgNO₃(aq)→ 2 NaNO₃(aq)+ Ag₂S(s) A)27.3 mL B)54.5 mL C)74.3 mL D)109 mL

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Quizplus · Accepted Answer

The balanced equation shows that 1 mole of Na2S reacts with 2 moles of AgNO3. Calculate moles of AgNO3: 0.045 L * 0.315 M = 0.014175 moles. Since the ratio is 1:2, moles of Na2S needed = 0.014175 / 2 = 0.0070875 moles. Volume of Na2S = moles / concentration = 0.0070875 / 0.260 = 0.0273 L = 27.3 mL.

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