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

There is 30 mL of CrCl3 necessary

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

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