A 55.0 mL solution of 0.550 M Pb(NO3)2 was mixed with 60.0 mL of 0.700 M Na3PO4, resulting in the reaction shown below: 3Pb(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 bo g 09

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A 55.0 mL solution of 0.550 M Pb(NO3)2 was mixed with 60.0 mL of 0.700 M Na3PO4, resulting in the
reaction shown below:
3Pb(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
09
09
g
Transcribed Image Text:A 55.0 mL solution of 0.550 M Pb(NO3)2 was mixed with 60.0 mL of 0.700 M Na3PO4, resulting in the reaction shown below: 3Pb(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 09 09 g
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