According to the following reaction, what volume in milliliters of 0.244M KCl (aq) solution is required to react exactly with 50.0 mL of 0.210 M Pb (NO3) 2?

2KCl (aq) + Pb (NO3) 2 (aq) - > PbCl2 (s) + 2KNO3 (aq)

There is 86.1 mL of KCl needed

Explanation:

Step 1: Data given

Molarity of KCl = 0.244 M

Volume of a 0.210 M Pb (NO3) 2 = 50.0 mL = 0.05 L

Step 2: The balanced equation

2KCl (aq) + Pb (NO3) 2 (aq) - > PbCl2 (s) + 2KNO3 (aq)

Step 3: Calculate moles Pb (NO3) 2

moles Pb (NO3) 2 = molarity * volume

moles Pb (NO3) 2 = 0.210 M * 0.05 L

moles Pb (NO3) 2 = 0.0105 moles

Step 4: Calculate moles of KCl

For 1 mole of Pb (NO3) 2 we need 2 moles of KCl

moles KCl required = 0.0105 * 2 = 0.0210 moles

Step 5: Calculate volume of KCl

Volume = moles KCl / molarity KCl

V = 0.0210 moles / 0.244 M=0.0861 L = 86.1 mL

There is 86.1 mL of KCl needed