Question

choose the correct products for the double replacement reaction below. click here to access the solubility rules to determine which product, if any, forms a solid precipitate in the reaction. pbcl₂ + mgso₄→? ○ pb + o₂ + mgcl ○ pbmg + clso₄ ○ pbso₄ + mgcl₂ ○ pb(so₄)₂ + mgcl

Explanation:

Step1: Recall Double Replacement Reaction

In a double replacement reaction, the cations and anions of the two reactants switch places. The general form is \( AB + CD
ightarrow AD + CB \), where \( A \) and \( C \) are cations, \( B \) and \( D \) are anions.

For the reaction \( \text{PbCl}_2 + \text{MgSO}_4
ightarrow \)?, the cation of the first reactant is \( \text{Pb}^{2+} \), anion is \( \text{Cl}^- \); the cation of the second reactant is \( \text{Mg}^{2+} \), anion is \( \text{SO}_4^{2-} \).

Step2: Swap Cations and Anions

Swapping the cations and anions, we get \( \text{Pb}^{2+} \) combining with \( \text{SO}_4^{2-} \) and \( \text{Mg}^{2+} \) combining with \( \text{Cl}^- \).

• The formula for the compound formed by \( \text{Pb}^{2+} \) and \( \text{SO}_4^{2-} \) is \( \text{PbSO}_4 \) (since the charges balance: \( 2+ + 2- = 0 \)).
• The formula for the compound formed by \( \text{Mg}^{2+} \) and \( \text{Cl}^- \) is \( \text{MgCl}_2 \) (since \( 2+ + 2\times(1-) = 0 \)).

So the products should be \( \text{PbSO}_4 + \text{MgCl}_2 \).

Answer:

\( \text{PbSO}_4 + \text{MgCl}_2 \) (corresponding to the option "PbSO₄ + MgCl₂")