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60.0 mL of 0.322 M potassium iodide are combined with 20.0 mL of 0.530 M lead (II) nitrate. How many grams of lead (II) iodide will precipitate? (you must write your own reaction)

Question

60.0 mL of 0.322 M potassium iodide are combined with 20.0 mL of 0.530 M lead (II) nitrate. 

How many grams of lead (II) iodide will precipitate? (you must write your own reaction)

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Step 1

The balanced equation for the reaction of potassium iodide with lead(II) nitrate is,

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Step 2

Find the limiting reagent using the mole ratio of the reactants. The reactants lead(II) nitrate and potassium iodide are in 1:2 mole ratio.

Determine the moles of each reactant as below.

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Step 3

Since, lead(II) nitrate is the limiting reactant, it means that all the moles of lead(II) nitrate are used by the reaction.  The number of moles of lead(II) iodide obtained as pro...

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