I. Solutions of potassium iodide and lead (II) nitrate combine to form a bright

Question

I. Solutions of potassium iodide and lead (II) nitrate combine to form a bright yellow precipitate.

a. How many grams of precipitate are formed if 3.45 grams of lead(II) nitrate is dissolved in water and added to a solution containing excess potassium iodide?


b. How many grams of precipitate are formed if 5.62 grams of potassium iodide is dissolved in water and added to a solution containing excess lead(II) nitrate?


c. How many grams of precipitate are formed if 8.62 grams of potassium iodide is dissolved in water and added to a solution containing 7.45 grams of lead(II) nitrate dissolved in water?

Explanation / Answer

The reaction that occurs is

2KI(aq) + Pb(NO3)2(aq) -> 2KNO3(aq) + PbI2(s)

a) Mol. wt of Pb(NO3)2 = 331.2

no. of moles= 3.45/331.2= 0.0104

So no. of moles of PbI2= 0.0104 because KI is in excess

wt of PbI= 0.0104*461.01 = 4.802gms

b) no. of moles of KI= 5.62/166.0028 = 0.034

no. of moles of PbI2= 0.034/2 = 0.017

wt. of PbI2 = 0.017*461.01= 7.83gms

c) no. of moles of KI= 8.62/166.0028= 0.0519

no. of moles of Pb(NO3)2= 7.45/331.2= 0.0225

0.0225< 0.0519/2 == 0.0225<0.02595

so no. of moles of PbI2 formed = 0.0225

wt of PbI2 = 0.0225*461.01 = 10.37gms

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