Lead II Nitrate & Barium Hydroxide

By - Timothy Sanya

About

Type of reaction: Double Replacement

Balanced Equation: Pb(NO3) (aq) + Ba(OH)2 (s) -------> Pb(OH)2 (s) + Ba(NO3)2 (aq)

IUPAC name: Lead(ll) Nitrate + Barium Hydroxide -----> Lead(ll) Hydroxide + Barium Nitrate

Molar Mass: 331.208 g/mol + 171.342 g/mol -------> 241.214 g/mol + 261.336 g/mol


Mole to mole conversions

6.04 moles Pb(NO3)2| 1 mole Ba(NO3)2

- | 1 mole Pb(NO3)2 = 6.04 moles Ba(NO3)2


Mass to mass conversions

12.1g Pb(NO3)2 |1 mole Pb(NO3)2 |1 mole Ba(OH)2 |171.342g Ba(OH)2

-| 331.208g Pb(NO3)2 |1 mole Pb(NO3)2 |1 mole Ba(OH)2 = 6.26g Ba(OH)2


Limiting and excess reactant

Limited- 12.3g Pb(NO3)2 |1 mole Pb(NO3)2 |1 mole Pb(OH)2 |241.214g Pb(OH)2

- | 331.208g Pb(NO3)2|1 mole Pb(NO3)2 | 1 mole Ba(OH)2 = 8.96g Pb(OH)2


Excess- 12.3g Ba(OH)2 | 1 mole Ba(OH)2 |1 mole Pb(OH)2 | 241.214g Pb(OH)2

- | 171.342g Ba(OH)2 |1 mole Ba(OH)2 | 1 mole Pb(OH)2 = 17.3g Pb(OH)2


Theoretical Yield: 8.96g Pb(OH)2


Percent Yield

11.02

8.96 x 100 = 123%

Lead II Nitrate & Barium Hydroxide