Question

when a lead nitrate solution is poured into a potassium iodide solution, a bright yellow solid called lead iodide forms. which is the balanced equation that represents this reaction?
○ a ( \text{pb(no}_3\text{)}_2 + \text{ki} longrightarrow \text{pbi}_2 + 2 \text{kno}_3 )
○ b ( \text{pb(no}_3\text{)}_2 + \text{ki} longrightarrow 2 \text{pbi}_2 + \text{kno}_3 )
○ c ( \text{pb(no}_3\text{)}_2 + 2 \text{ki} longrightarrow \text{pbi}_2 + \text{kno}_3 )
○ d ( \text{pb(no}_3\text{)}_2 + 2 \text{ki} longrightarrow \text{pbi}_2 + 2 \text{kno}_3 )

Explanation:

Step1: Analyze Lead (Pb) atoms

On the left side, we have 1 Pb atom from $\ce{Pb(NO_{3})_{2}}$. On the right side, in $\ce{PbI_{2}}$, we also have 1 Pb atom. So Pb is balanced for now.

Step2: Analyze Nitrate ($\ce{NO_{3}^-}$) ions

Left side: 2 $\ce{NO_{3}^-}$ ions from $\ce{Pb(NO_{3})_{2}}$. Right side: In $\ce{KNO_{3}}$, we need 2 $\ce{NO_{3}^-}$ ions to balance, so we need 2 $\ce{KNO_{3}}$.

Step3: Analyze Potassium (K) atoms

Right side: 2 K atoms from 2 $\ce{KNO_{3}}$. So left side needs 2 K atoms, which means 2 $\ce{KI}$ (since each $\ce{KI}$ has 1 K atom).

Step4: Analyze Iodine (I) atoms

Left side: 2 I atoms from 2 $\ce{KI}$. Right side: In $\ce{PbI_{2}}$, we have 2 I atoms. So I is balanced.

Now let's check each option:

• Option a: $\ce{Pb(NO_{3})_{2} + KI -> PbI_{2} + 2KNO_{3}}$

K: Left has 1, right has 2. Not balanced. I: Left has 1, right has 2. Not balanced.

• Option b: $\ce{Pb(NO_{3})_{2} + KI -> 2PbI_{2} + KNO_{3}}$

Pb: Left has 1, right has 2. Not balanced.

• Option c: $\ce{Pb(NO_{3})_{2} + 2KI -> PbI_{2} + KNO_{3}}$

$\ce{NO_{3}^-}$: Left has 2, right has 1. Not balanced. K: Left has 2, right has 1. Not balanced.

• Option d: $\ce{Pb(NO_{3})_{2} + 2KI -> PbI_{2} + 2KNO_{3}}$

Pb: 1 on both sides. $\ce{NO_{3}^-}$: 2 on both sides. K: 2 on both sides. I: 2 on both sides. Balanced.

Answer:

d. $\ce{Pb(NO_{3})_{2} + 2 KI -> PbI_{2} + 2 KNO_{3}}$