Answered

Westonci.ca is the best place to get answers to your questions, provided by a community of experienced and knowledgeable experts. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.

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.

Sagot :

ardni313

The mass of the Pbl2 : 1308.87

Further explanation

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 :

[tex]\tt mass=mol\times MW\\\\mass=2.8392\times 461\\\\mass=1308.87~grams[/tex]

We hope this information was helpful. Feel free to return anytime for more answers to your questions and concerns. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.