Question

1. balance the double replacement reaction by dragging the correct coefficients.

? pb(no₃)₂(aq) + ? ki(aq) → ? pbi₂(s) + ? kno₃(aq)
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Explanation:

Step1: Balance Pb atoms

There is 1 Pb atom on the left (in \( \text{Pb(NO}_3\text{)}_2 \)) and 1 Pb atom on the right (in \( \text{PbI}_2 \)), so Pb is balanced initially.

Step2: Balance I atoms

On the right, \( \text{PbI}_2 \) has 2 I atoms. On the left, KI has 1 I atom. So we need a coefficient of 2 for KI to balance I: \( 2\text{KI} \).

Step3: Balance K and \( \text{NO}_3 \)

After balancing I with 2 KI, we have 2 K atoms on the left. So we need 2 \( \text{KNO}_3 \) on the right to balance K. Now, checking \( \text{NO}_3 \): \( \text{Pb(NO}_3\text{)}_2 \) has 2 \( \text{NO}_3^- \) groups, and \( 2\text{KNO}_3 \) has 2 \( \text{NO}_3^- \) groups, so \( \text{NO}_3 \) is balanced.

The balanced equation is: \( \text{Pb(NO}_3\text{)}_2 + 2\text{KI}
ightarrow \text{PbI}_2 + 2\text{KNO}_3 \)

Answer:

The coefficients are: \( \text{Pb(NO}_3\text{)}_2 \): 1, \( \text{KI} \): 2, \( \text{PbI}_2 \): 1, \( \text{KNO}_3 \): 2. So the balanced equation is \( 1\text{Pb(NO}_3\text{)}_2 + 2\text{KI}
ightarrow 1\text{PbI}_2 + 2\text{KNO}_3 \)