A 70.0 mL solution of 0.550 M Pb(NO3)2 was mixed with 55.0 mL of 0.600 M Na3PO4, resulting in the reaction shown below: 3 Pb(NO3)2 (aq) + 2 Na3PO4 (aq) → Pb3(PO4)2 (s) + 6 NaNO3(aq) a) What are the molecular weights of Pb(NO3)2, Na3PO4, Pb3(PO4)2? Pb(NO3)2: Na3PO4: Pb3(PO4)2: b) Determine the theoretical amount of Pb3(PO4)2 (s) formed in grams. Mass: c) Determine the amount of unused reactant in grams. Mass: g/mol g/mol g/mol 8.0 g 80 g

Chemistry: Principles and Practice
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ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Chapter4: Chemical Reactions In Solution
Section: Chapter Questions
Problem 4.120QE
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A 70.0 mL solution of 0.550 M Pb(NO3)2 was mixed with 55.0 mL of 0.600 M Na3PO4, resulting in the
reaction shown below:
a) What are the molecular weights of Pb(NO3)2, Na3PO4, Pb3(PO4)2?
3 Pb(NO3)2 (aq) + 2 Na3PO4 (aq) → Pb3(PO4)2 (s) + 6 NaNO3 (aq)
Pb(NO3)2 :
Na3PO4:
Pb3(PO4)2
b) Determine the theoretical amount of Pb3(PO4)2 (s) formed in grams.
Mass:
c) Determine the amount of unused reactant in grams.
Mass:
g/mol
g/mol
g/mol
80
g
g
مة
Transcribed Image Text:A 70.0 mL solution of 0.550 M Pb(NO3)2 was mixed with 55.0 mL of 0.600 M Na3PO4, resulting in the reaction shown below: a) What are the molecular weights of Pb(NO3)2, Na3PO4, Pb3(PO4)2? 3 Pb(NO3)2 (aq) + 2 Na3PO4 (aq) → Pb3(PO4)2 (s) + 6 NaNO3 (aq) Pb(NO3)2 : Na3PO4: Pb3(PO4)2 b) Determine the theoretical amount of Pb3(PO4)2 (s) formed in grams. Mass: c) Determine the amount of unused reactant in grams. Mass: g/mol g/mol g/mol 80 g g مة
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