Question

In the reaction between sodium iodide (NaI) and lead(II) nitrate (Pb(NO₃)₂, which produces lead(II) iodide precipitation and sodium nitrate aqueous solution, what is the balanced chemical equation?

A. 2 NaI(aq) + Pb(NO₃)₂(aq) → 2 NaNO₃(aq) + PbI₂(s)
B. NaI(aq) + Pb(NO₃)₂(aq) → NaNO₃(aq) + PbI₂(s)
C. 3 NaI(aq) + 2 Pb(NO₃)₂(aq) → 2 NaNO₃(aq) + PbI₂(s)
D. 2 NaI(aq) + Pb(NO₃)₂(aq) → NaNO₃(aq) + PbI₂(s)

Answer 1

The balanced chemical equation for the reaction between sodium iodide and lead(II) nitrate is 2 NaI(aq) + Pb(NO₃)₂(aq) → 2 NaNO₃(aq) + PbI₂(s), which is a double displacement reaction resulting in lead(II) iodide precipitation.

Step-by-step explanation:

The balanced chemical equation for the reaction between sodium iodide (NaI) and lead(II) nitrate (Pb(NO₃)₂) that produces lead(II) iodide precipitation and a sodium nitrate aqueous solution is represented as:

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

This reaction is a precipitation reaction where lead(II) iodide forms as a solid precipitate, and is confirmed by the given source as well. It follows the pattern of a double displacement or metathesis reaction, where the anions and cations of two different compounds exchange places and form two new compounds, one of which is usually a precipitate.