. A reaction of 220. mL of a lead (II) nitrate solution with excess hydroiodic acid produced 5.50 grams of solid lead (II) iodide and nitric acid. a. Determine the moles of lead (II) iodide produced b. Determine the moles of lead (II) nitrate that would react with the moles of lead (II) iodide calculated in part a.

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. A reaction of 220. mL of a lead (II)
nitrate solution with excess hydroiodic
acid produced 5.50 grams of solid lead
(II) iodide and nitric acid. a. Determine
the moles of lead (II) iodide produced b.
Determine the moles of lead (II) nitrate
that would react with the moles of lead
(II) iodide calculated in part a.
Transcribed Image Text:. A reaction of 220. mL of a lead (II) nitrate solution with excess hydroiodic acid produced 5.50 grams of solid lead (II) iodide and nitric acid. a. Determine the moles of lead (II) iodide produced b. Determine the moles of lead (II) nitrate that would react with the moles of lead (II) iodide calculated in part a.
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