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.
1 answer:
The mass of the Pbl2 : 1308.87
<h3>Further explanation</h3>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 :
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The attached diagrams portraying this notions is shown in the attached file below.
