18.2 mL of a 0.156 M solution of lead(II) nitrate are added to 26.2 mL of a 0.274 M solution of potassium iodide. What is the mass of the Pbl2 precipitate formed in the reaction Pb(NO3)2 (aq) + 2 KI (aq) - Pbl2 (s) + 2 KNO3? The molar mass of Pbly is 461.0 g/mol.
Provide your answer in units of grams, with the correct number of significant digits. Enter your answer as a number only; do not include units.

The mass of the Pbl2 : 1308.87

Given

18.2 mL of a 0.156 M Pb(NO3)2

26.2 mL of a 0.274 M KI

Reaction

Pb(NO3)2 (aq) + 2 KI (aq) - Pbl2 (s) + 2 KNO3

Required

the mass of the Pbl2

Solution

mol Pb(NO3)2 = 18.2 x 0.156 = 2.8392 mlmol

mol KI = 26.2 x 0.274 =7.1788 mol

Limiting reactant Pb(NO3)2(smaller ratio of mol : reaction coeffiecient)

mol Pbl2 based on limiting reactant (Pb(NO3)2)

From equation, mol ratio of Pb(NO3)2 : Pbl2 = 1 : 1, so mol Pbl2=mol Pb(NO3)2=2.8392

Mass Pb(NO3)2 :

it's filtration, but this is an assumption so my apologies if i got this

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