What mass of precipitate will form if 1.50 L of concentrated Pb(ClO3)2 is mixed with 0.700 L of 0.240 M NaI? Assume the reaction goes to completion.

Answer 1

Assuming NaI is the limiting reactant,
so mass of PbI2 ..
moles NaI=.7*.240=0.168moles, which means .168*2 moles of PbI2, so convert that to mass.
mass=.168*2*(461) grams.