Answer:
1.24g of precipitate are produced
Step-by-step explanation:
The reaction of NaCl with Pb(NO₃)₂ is:
2 NaCl + Pb(NO₃)₂ → 2 NaNO₃ + PbCl₂(s)
Where 2 moles of NaCl reacts with 1 mole of Pb(NO₃)₂ to produce 1 mole of PbCl₂, the precipitate.
To know how much PbCl₂ is formed we need to determine the moles of NaCl and Pb(NO₃)₂:
Moles NaCl:
35.0mL = 0.0350L * (0.255mol / L) = 8.925x10⁻³ moles NaCl
Moles Pb(NO₃)₂:
45.0mL = 0.0450L * (0.328mol / L) = 0.01476 moles Pb(NO₃)₂
For a complete reaction of 0.01476 moles Pb(NO₃)₂, you require:
0.01476 moles Pb(NO₃)₂ * (2 moles NaCl / 1 mol Pb(NO₃)₂) = 0.02952 moles NaCl.
As you have just 8.925x10⁻³ moles, NaCl is limiting reactant.
And moles of PbCl₂ produced are:
8.925x10⁻³ moles NaCl * (1mol PbCl₂ / 2 moles NaCl) =
4.4625x10⁻³ moles PbCl₂ are formed.
In grams, (Using molar mass PbCl₂: 278.1g/mol):
4.4625x10⁻³ moles PbCl₂ * (278.1g / mol) =
1.24g of precipitate are produced