27.45mLs of 0.250M lead (II) nitrate is reacted with 22.76mLs of 0.450M sodium iodide. How many grams of lead (II) iodide precipitate? (aq) + (aq) → Name Lead (II) nitrate sodium iodide Measurement Conversion moles Calculations 0.450mol sodium iodide 1 L solution (s) + lead (II) iodide not applicable 461.01g not applicable 1 mol not applicable (aq)

Chemistry: Principles and Reactions
8th Edition
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:William L. Masterton, Cecile N. Hurley
Chapter4: Reactions In Aqueous Solution
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Problem 9QAP: Write a net ionic equation for any precipitation reaction that occurs when 1 M solutions of the...
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Determine how many moles of sodium iodide were consumed and how many moles of lead (II) iodide were produced?

27.45mLs of 0.250M lead (II) nitrate is reacted with 22.76mLs of 0.450M sodium iodide. How many grams of lead
(II) iodide precipitate?
(aq) +
(aq)
→
(s) +
Name
Lead (II) nitrate
sodium iodide
lead (II) iodide
not applicable
Measurement
0.450mol sodium iodide
461.01g
Conversion
1 L solution
1 mol
not applicable
moles
Calculations
not applicable
(aq)
Transcribed Image Text:27.45mLs of 0.250M lead (II) nitrate is reacted with 22.76mLs of 0.450M sodium iodide. How many grams of lead (II) iodide precipitate? (aq) + (aq) → (s) + Name Lead (II) nitrate sodium iodide lead (II) iodide not applicable Measurement 0.450mol sodium iodide 461.01g Conversion 1 L solution 1 mol not applicable moles Calculations not applicable (aq)
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