How many milliliters of 0.260 M Na₂S are needed to react with 40.00 mL of 0.315 M AgNO₃?
Na₂S(aq) + 2 AgNO₃(aq) → 2 NaNO₃(aq) + Ag₂S(s)

Question

How many milliliters of 0.260 M Na₂S are needed to react with 40.00 mL of 0.315 M AgNO₃? Na₂S(aq)+ 2 AgNO₃(aq)→ 2 NaNO₃(aq)+ Ag₂S(s)

Quizplus · Accepted Answer

The balanced chemical equation is:Na₂S(aq)+ 2 AgNO₃(aq)→ 2 NaNO₃(aq)+ Ag₂S(s)From the balanced chemical equation, we can see that 1 mole of Na₂S reacts with 2 moles of AgNO₃. Therefore, the number of moles of Na₂S needed to react with 40.00 mL of 0.315 M AgNO₃ is:(40.00 mL)(0.315 M) / 1000 mL/L = 0.0126 mol AgNO₃(0.0126 mol AgNO₃) * (1 mol Na₂S / 2 mol AgNO₃) = 0.0063 mol Na₂SThe volume of 0.260 M Na₂S needed to provide 0.0063 mol of Na₂S is:(0.0063 mol Na₂S) * (1 L / 0.260 mol Na₂S) = 0.0242 L = 24.2 mLTherefore, the correct answer is A) 24.2 mL.

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