Question

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)

Answer 1

There is 86.1 mL of KCl needed

Step-by-step 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