Question

If 20.0 mL of 0.400 M Pb(NO₃)₂ and 60.0 mL of 0.200 M NaCl are mixed, what mass (in grams) of PbCl₂precipitate will form?

(Hint: FW of PbCl₂ = 278.11 g/mol)

Answer 1

the balanced equation is as follows

Pb(NO3)2 + 2 NaCl ------------------------> PbCl2 + 2 NaNO3

no of moles = mass / molar mass

for Pb(NO3)2 = 0.02 L * 0.40 M = 0.008 moles

for NaCl = 0.06 L * 0.200 M = 0.012 moles

from the balanced equation

=> 1 mol Pb(NO3)2 ---------------------------> 1 mol PbCl2

     0.008 moles --------------------------------> 0.008 moles

=> 2 mole NaCl -------------------------> 1 mol PbCl2

      0.012 moles ------------------------->?

    => 0.012*1/2 = 0.06 moles

so the limitting reactant is PbNO32

theoretical yield of PbCl2 = > moles * molar mass = 0.008 moles * 278.11 g/mol

                                            => 2.224 grams