Write the balanced equation for the reaction of aqueous Pb (ClO 3) 2 with aqueous NaI. Include phases.

The balanced equation for the reaction is:

Pb (ClO₃) ₂ (aq) + 2NaI (aq) → PbI₂ (s) ↓ + 2NaClO₃

Explanation:

The reaction of Pb²⁺ with I⁻ makes a precipitate of lead iodide

Pb (ClO3) 2 (aq) + 2NaI (aq) → PbI2 (s) + 2Na (ClO3) (aq)

Explanation:

Step 1: Data given

Reactants are: Pb (ClO3) 2 and NaI

Step 2: Balancing the equation

Pb (ClO3) 2 (aq) + NaI (aq) → PbI (s) + NaClO3 (aq)

On the left side, we have 2x Cl, on the right side we have 1x Cl

To balance the amount of Cl, we have to multiply Na (ClO3) 3 by 2

Pb (ClO3) 2 (aq) + NaI (aq) → PbI2 (s) + 2Na (ClO3) (aq)

On the right side we have 2x Na, on the left side we have 1x Na.

To balance the amount of Na, we have to multiply NaI (on the left side) by 2

Now the equation is balanced

Pb (ClO3) 2 (aq) + 2NaI (aq) → PbI2 (s) + 2Na (ClO3) (aq)