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 0.300 M NaI

The mass of precipitate (lead (II) iodide) that will form is 13.83 grams

Explanation:

To calculate the number of moles for given molarity, we use the equation:

Molarity of NaI solution = 0.300 M

Volume of solution = 0.200 L

Putting values in above equation, we get:

The balanced chemical equation for the reaction of lead chlorate and sodium iodide follows:

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, 0.06 moles of NaI will produce = of lead (II) iodide

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

Moles of lead (II) iodide = 0.03 moles

Molar mass of lead (II) iodide = 461.1 g/mol

Putting values in above equation, we get:

Hence, the mass of precipitate (lead (II) iodide) that will form is 13.83 grams

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