According to the following reaction,what volume of 0.244 M KCl solution is required to react exactly with 50.0 mL of 0.210 M Pb(NO₃)₂ solution? 2 KCl(aq)+ Pb(NO₃)₂(aq)→ PbCl₂(s)+ 2 KNO₃(aq) A) 97.4 mL B) 116 mL C) 43.0 mL D) 86.1 mL E) 58.1 mL

Question

Quizplus · Accepted Answer

First, we need to use stoichiometry to find the moles of Pb(NO₃)₂ that will react with the KCl solution. 50.0 mL of 0.210 M Pb(NO₃)₂ solution contains: 0.0500 L x 0.210 mol/L = 0.0105 mol of Pb(NO₃)₂ From the balanced equation, we see that 2 moles of KCl react with 1 mole of Pb(NO₃)₂. Therefore, we need: 2 mol KCl / 1 mol Pb(NO₃)₂ x 0.0105 mol Pb(NO₃)₂ = 0.0210 mol KCl Now we can use the molarity and volume of the KCl solution to find the volume required: 0.244 M KCl = 0.244 mol/L 0.0210 mol / 0.244 mol/L = 0.0861 L = 86.1 mL Therefore, the answer is D) 86.1 mL.

According to the Following Reaction,what Volume of 0