Question

How many mL of 1.0M CrCl3 solution would be necessary to precipitate all of the Pb2+ from 50mL of .90M solution.2CrCl3 + 3Pb(NO3)2 ---> 3PbCl2 + 2Cr(NO3)3A.) 10B.) 20C.) 30D.) 60

Answer 1

There is 30 mL of CrCl3 necessary

Step-by-step explanation:

Step 1: Data given

Volume of Pb2+ = 50 mL = 0.05L

Molarity of Pb2+ = 0.90 M

Molarity of CrCl3 = 1.0 M

Step 2: The balanced equation

2CrCl3 + 3Pb(NO3)2 → 3PbCl2 + 2Cr(NO3)3

3PbCl2 → 3Pb2+ 6Cl-

For 2 moles CrCl3 consumed, we need 3 moles Pb(NO3)2 to produce 3 moles of PbCl2 and 2 moles of Cr(NO3)3

Step 3: Calculate moles Pb2+

Moles Pb2+ = Molarity Pb2+ * volume

Moles Pb2+ = 0.90 M * 0.05 L

Moles Pb2+ = 0.045 moles

Step 4: Calculate moles PbCl2

For 2 moles CrCl3 consumed, we need 3 moles Pb(NO3)2 to produce 3 moles of PbCl2 and 2 moles of Cr(NO3)3

For 0.045 moles of pb2+ we have 0.045 moles of PbCl2

Step 5: Calculate moles of CrCl3

For 3 moles PbCl2 produced, we need 2 moles CrCl3 consumed.

For 0.045 moles of PbCl2, we need 0.030 moles of CrCl3

Step 6: Calculate volume of CrCl3 solution

Volume = moles CrCl3 /Molarity

volume = 0.030 moles / 1M

volume = 0.030 L = 30mL

There is 30 mL of CrCl3 necessary