Question

Potassium iodide, Kl, reacts with lead(II) nitrate according to the following chemical equation. 

2 Kl (aq) + Pb(NO₃)₂ (aq) → 2 KNO₃ (aq) + Pbl₂ (s) 

If 32.31 mL of 0.383 M Kl reacts with excess Pb(NO₃)₂ , what mass of Pbl forms? The molar mass of Pbl₂ is 461.01 g/mol.

Answer 1

32.31 mL of 0.383 M KI react with Pb(NO₃)2

Number of moles = Volume in L × concentration

1L = 1000 ml , 1ml = 0.001L

Number of moles = 32.31 × 10^-3 × 0.383 = 0.012 mol

Now given equation

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

Since Pb(NO3)2 is given in excess

Therefore KI is limiting reagent.

So 2 moles of KI produce = 1 mole of PbI2

1 mole of KI produce = 1/2 moles of PbI2

Therefore 0.012 moles of KI produce = 0.012 × 0.5 = 0.006 moles of PbI2

Now Number of moles of PbI2 produce = mass of PbI2 / Molecular weight of PbI2

=> 0.006 mole = mass of PbI2 / 461.01 g/mole

Mass of PbI2 produce = 0.006 × 461.01 = 2.7 g

2.7 grams of PbI2 produce.

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