Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 Pb(ClO3)2 with aqueous NaI . NaI. Include phases. chemical equation: What mass of precipitate will form if 1.50 L 1.50 L of highly concentrated Pb ( ClO 3 ) 2 Pb(ClO3)2 is mixed with 0.400 L 0.130 M NaI 0.400 L 0.130 M NaI ? Assume the reaction goes to completion

The mass of precipitate (lead (II) iodide) that will form is 119.89 grams

Explanation:

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

thats a little if u want the answer just look it up on google thats what i do haha : )