Answer:
[tex]m_{PbI_2}=23.6gPbI_2[/tex]
Explanation:
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In this case, since the undergoing chemical reaction is:
[tex]2KI+Pb(NO_3)_2\rightarrow 2KNO_3+PbI_2[/tex]
Thus, given the masses of potassium iodide (166.0028 g/mol) and lead (II) nitrate (331.2 g/mol), we compute the yielded moles of lead (II) iodide in order to identify the limiting reactant:
[tex]n_{PbI_2}^{by\ KI}=17.0gKI*\frac{1molKI}{166.0gKI}*\frac{1molPbI_2}{2molKI} =0.0512molPbI_2\\\\n_{PbI_2}^{by\ Pb(NO_3)_2}=25gPb(NO_3)_2*\frac{1molPb(NO_3)_2}{331.2gPb(NO_3)_2}*\frac{1molPbI_2}{1molPb(NO_3)_2} =0.0755molPbI_2[/tex]
Thus, since potassium iodide yields the fewest moles of product, we infer that 0.0512 moles of lead (II) iodide are theoretically yielded, so the associated mass is:
[tex]m_{PbI_2}=0.0512molPbI_2*\frac{461.01gPbI_2}{1molPbI_2} \\\\m_{PbI_2}=23.6gPbI_2[/tex]
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