Question

Lead Nitrate react with potassium iodide to form a precipitate of lead iodide and potassium nitrate

Answer 1

Here is the reaction...

Answer 2

"It seems this is what you are looking for"

Lead (II) nitrate reacts with potassium iodide to form lead (II) iodide and potassium nitrate. How do you write the balanced equation for this?

Answer:

The precipitation reaction of lead (II) nitrate reacts with potassium iodide will produce lead (II) iodide and potassium nitrate.

The balanced chemical equation is:

      $Pb(NO_{3})_{2} + 2KI$   →  $PbI_{2} + 2KNO_{3}$

Explanation:

The unbalanced chemical equation of given precipitation reaction:

       $Pb(NO_{3})_{2} + KI$  →   $PbI_{2} + KNO_{3}$

Make a list of number of atoms in the unbalanced chemical equation

              $Atom\\ \\Pb\\I\\N\\O\\K$                    $Reactant\\side\\1\\1\\2\\6\\1$            $Product\\side\\1\\2\\1\\3\\1$

• Multiple with coefficients to balance the mass on both sides.
• Try to balance the single atom like iodine first then move to nitrogen or oxygen.
• To balance the iodine on both sides multiple KI with 2.
• Then to balance  K multiple KNO₃ with 2.
• Nitrogen and oxygen will automatically get balanced.

In this way, we will get a balanced chemical equation:

             $Pb(NO_{3})_{2} + 2KI$    →    $PbI_{2} + 2KNO_{3}$