Question

question 1 of 27
the balanced equation for the reaction of aqueous ( mathrm{pb}left(mathrm{clo}_{3}
ight)_{2} ) with aqueous ( mathrm{nai} ) is
mathrm{pb}left(mathrm{clo}_{3}
ight)_{2}(mathrm{aq})+2 mathrm{nai}(mathrm{aq}) \to mathrm{pbl}_{2}(mathrm{~s})+2 mathrm{naclo}_{3}(mathrm{aq})
what mass of precipitate will form if ( 1.50 mathrm{~l} ) of concentrated ( mathrm{pb}left(mathrm{clo}_{3}
ight)_{2} ) is mixed with ( 0.450 mathrm{~l} ) of ( 0.260 mathrm{m} mathrm{nai} ) ? assume the
reaction goes to completion.
mass of precipitate: ( mathrm{g} )

Explanation:

Step1: Calculate the moles of NaI

Use the formula \(n = M\times V\). Given \(M = 0.260\space M\) and \(V=0.450\space L\).
\(n_{NaI}=0.260\space M\times0.450\space L = 0.117\space mol\)

Step2: Determine the mole ratio from the balanced equation

From \(Pb(ClO_3)_2(aq)+2NaI(aq)\to PbI_2(s)+2NaClO_3(aq)\), the mole ratio of \(NaI\) to \(PbI_2\) is \(2:1\).

Step3: Calculate the moles of \(PbI_2\)

Let \(n_{PbI_2}\) be the moles of \(PbI_2\). Using the mole ratio \(\frac{n_{PbI_2}}{n_{NaI}}=\frac{1}{2}\), so \(n_{PbI_2}=\frac{0.117\space mol}{2}=0.0585\space mol\)

Step4: Calculate the molar mass of \(PbI_2\)

The molar mass of \(Pb = 207.2\space g/mol\) and \(I = 126.9\space g/mol\). So \(M_{PbI_2}=207.2+(2\times126.9)= 461\space g/mol\)

Step5: Calculate the mass of \(PbI_2\)

Use the formula \(m = n\times M\). Substitute \(n = 0.0585\space mol\) and \(M = 461\space g/mol\)
\(m_{PbI_2}=0.0585\space mol\times461\space g/mol = 27.0\space g\)

Answer:

\(27.0\space g\)