Alright, let's classify the given chemical reaction:
[tex]\[ Pb(NO_3)_2 + Na_2CrO_4 \rightarrow PbCrO_4 + 2 NaNO_3 \][/tex]
### Step-by-Step Solution:
1. Identify the Reactants and Products:
- Reactants: Lead(II) nitrate ([tex]\(Pb(NO_3)_2\)[/tex]) and sodium chromate ([tex]\(Na_2CrO_4\)[/tex]).
- Products: Lead(II) chromate ([tex]\(PbCrO_4\)[/tex]) and sodium nitrate ([tex]\(NaNO_3\)[/tex]).
2. Analyze the Reaction Type:
- Notice that in the reactants, we have two compounds:
- Lead(II) nitrate ([tex]\(Pb(NO_3)_2\)[/tex])
- Sodium chromate ([tex]\(Na_2CrO_4\)[/tex])
- In the products, we also have two compounds:
- Lead(II) chromate ([tex]\(PbCrO_4\)[/tex])
- Sodium nitrate ([tex]\(NaNO_3\)[/tex])
3. Check for Ion Exchange:
- In the reaction, [tex]\(Pb^{2+}\)[/tex] (a cation) from [tex]\(Pb(NO_3)_2\)[/tex] pairs with [tex]\(CrO_4^{2-}\)[/tex] (an anion) from [tex]\(Na_2CrO_4\)[/tex] to form [tex]\(PbCrO_4\)[/tex].
- Similarly, [tex]\(Na^+\)[/tex] (a cation) pairs with [tex]\(NO_3^-\)[/tex] (an anion) to form [tex]\(NaNO_3\)[/tex].
4. Determine the Type of Reaction:
- Here, the cations (Pb[tex]\(^{2+}\)[/tex] and Na[tex]\(^{+}\)[/tex]) and the anions (NO[tex]\(_3^-\)[/tex] and CrO[tex]\(_4^{2-}\)[/tex]) essentially switch partners to form two new compounds: [tex]\(PbCrO_4\)[/tex] and [tex]\(NaNO_3\)[/tex].
5. Classify the Reaction:
- This type of reaction, where two compounds exchange partners to form two new compounds, is known as a double replacement reaction.
### Conclusion:
- The given reaction [tex]\( Pb(NO_3)_2 + Na_2CrO_4 \rightarrow PbCrO_4 + 2 NaNO_3 \)[/tex] is best classified as a double replacement reaction.
