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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? The unbalanced equation of reaction is:

Pb(NO3)2(aq) + NaIO3(aq) → Pb(IO3)2(s) + NaNO3(aq)

  • chemistry -

    Balance the equation.
    Calculate moles of each from moles = M x L.
    Determine the limiting reagent.
    moles Pb(IO3)2 x molar mass = grams.

    I note there is an excess of NaIO3 present which will reduce the solubility but I don't think you need to consider that.

  • chemistry -

    24.4 g

  • chemistry -


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