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<u>Answer:</u> The mass of precipitate (lead (II) iodide) that will form is 119.89 grams
<u>Explanation:</u>
To calculate the number of moles for given molarity, we use the equation:
Molarity of NaI solution = 0.130 M
Volume of solution = 0.400 L
Putting values in above equation, we get:
The balanced chemical equation for the reaction of lead chlorate and sodium iodide follows:
The precipitate (insoluble salt) formed is lead (II) iodide
By Stoichiometry of the reaction:
2 moles of NaI produces 1 mole of lead (II) iodide
So, 0.52 moles of NaI will produce = of lead (II) iodide
To calculate the number of moles, we use the equation:
Moles of lead (II) iodide = 0.26 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 119.89 grams