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How Many Grams of Lead(II)iodate, Pb(IO3)2 (Formula Weight = 557

Question 234

Multiple Choice

How many grams of lead(II) iodate, Pb(IO3) 2 (formula weight = 557.0 g/mol) , are precipitated when 3.20 × 102 mL of 0.285 M Pb(NO3) 2(aq) are mixed with 386 mL of 0.512 M NaIO3(aq) solution? Hint: Write out the balanced equation and remember to use the correct units when dealing with molarity.


A) 25.4 g
B) 39.8 g
C) 48.3 g
D) 50.8 g
E) 55.0 g

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