Question

a solution of lead (ii) nitrate reacts with a solution of sodium iodide to produce a yellow solid of lead (ii) iodide and aqueous sodium nitrate. which of these illustrates the reaction?
○ ( mathrm{pb}left(mathrm{no}_{3}
ight)_{2}(mathrm{aq})+2 mathrm{nal}(mathrm{aq})
ightarrow mathrm{pbi}_{2}(mathrm{~s})+2 mathrm{nano}_{3}(mathrm{aq}) )
○ ( mathrm{l}_{2} mathrm{~n}(mathrm{aq})+mathrm{si}(mathrm{aq})
ightarrow mathrm{l}_{2} mathrm{i}(mathrm{s})+mathrm{sn}(mathrm{aq}) )
○ ( mathrm{pb}_{2} mathrm{~n}(mathrm{aq})+mathrm{nal}(mathrm{aq})
ightarrow mathrm{pb}_{2} mathrm{i}(mathrm{s})+mathrm{nan}(mathrm{aq}) )
○ ( mathrm{fe}left(mathrm{no}_{3}
ight)_{2}(mathrm{aq})+2 mathrm{nal}(mathrm{aq})
ightarrow mathrm{fel}_{2}+2 mathrm{nano}_{3}(mathrm{aq}) )

Explanation:

Step1: Analyze reactants

Lead (II) nitrate is Pb(NO₃)₂, sodium iodide is NaI.

Step2: Analyze products

Lead (II) iodide is PbI₂ (solid, s), sodium nitrate is NaNO₃ (aqueous, aq).

Step3: Balance the equation

For nitrate (NO₃⁻) and iodide (I⁻) ions, we have 2NaI to balance the equation.

Answer:

Pb(NO₃)₂(aq) + 2NaI(aq) → PbI₂(s) + 2NaNO₃(aq)