How many milliliters of 0.300 M Li₂S are needed to react with 25.00 mL of 0.315 M AgNO₃? Li₂S(aq)+ 2 AgNO₃(aq)→ 2 LiNO₃(aq)+ Ag₂S(s) A)13.1 mL B)226.2 mL C)6.05 mL D)1.31 mL

Question

Quizplus · Accepted Answer

The balanced chemical equation is:Li₂S(aq)+ 2 AgNO₃(aq)→ 2 LiNO₃(aq)+ Ag₂S(s)From the balanced chemical equation, we can see that 1 mole of Li₂S reacts with 2 moles of AgNO₃. Therefore, the number of moles of Li₂S needed to react with 25.00 mL of 0.315 M AgNO₃ is:(25.00 mL)(0.315 M) / 1000 mL/L = 0.007875 mol AgNO₃(0.007875 mol AgNO₃)(1 mol Li₂S / 2 mol AgNO₃) = 0.003938 mol Li₂SThe volume of 0.300 M Li₂S needed to provide 0.003938 mol of Li₂S is:(0.003938 mol Li₂S)(1 L / 0.300 M) = 0.0131 mLTherefore, the correct answer is A) 13.1 mL.

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