How many grams of Pb(IO3)2 are precipitated when 3.20´102 mL of 0.285 M Pb(NO3)2(aq) are mixed with 325 mL of 0.512 M NaIO3(aq) solution?
The reaction is: Pb(NO3)2(aq) + NaIO3(aq) -- > Pb(IO3)2(s) + NaNO3(aq)
(Hint: balanced equation? limiting reactant?)
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I don't know.