Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 with aqueous NaI . Include phases. chemical equation: What mass of precipitate will form if 1.50 L of highly concentrated Pb ( ClO 3 ) 2 is mixed with 0.200 L 0.200 M NaI ? Assume the reaction goes to completion.

The mass of lead iodide produced is 9.22 grams

Explanation:

To calculate the molarity of solution, we use the equation:

Molarity of NaI = 0.200 M

Volume of solution = 0.200 L

Putting values in above equation, we get:

The chemical equation for the reaction of NaI and lead chlorate follows:

By Stoichiometry of the reaction:

2 moles of NaI reacts produces 1 mole of lead iodide

So, 0.04 moles of NaI will react with = of lead iodide

To calculate the number of moles, we use the equation:

Molar mass of lead iodide = 461 g/mol

Moles of lead iodide= 0.02 moles

Putting values in above equation, we get:

Hence, the mass of lead iodide produced is 9.22 grams

dynamic equilibrium is the answer.