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.

Question

Quizplus · Accepted Answer

The balanced equation is Pb(NO3)2 + 2NaIO3 → Pb(IO3)2 + 2NaNO3. Calculate moles of Pb(NO3)2: 0.285 mol/L * 0.320 L = 0.0912 mol. Calculate moles of NaIO3: 0.512 mol/L * 0.386 L = 0.1978 mol. Pb(IO3)2 requires 2 moles of NaIO3 per mole of Pb(NO3)2, so Pb(NO3)2 is the limiting reagent. Moles of Pb(IO3)2 formed = 0.0912 mol. Mass = 0.0912 mol * 557.0 g/mol = 50.8 g.

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

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