Respuesta :
The balanced chemical equation for the reaction
Pb(ClO₃)₂(aq) + 2NaI(aq) → PbI₂(s) + 2NaClO₃(aq)
The mass of precipitate that would form is 11.986 g
From the question,
We are to write the balanced chemical equation for the reaction, of aqueous Pb(ClO₃)₂ with aqueous NaI
The balanced chemical equation for the reaction
Pb(ClO₃)₂(aq) + 2NaI(aq) → PbI₂(s) + 2NaClO₃(aq)
Now, we are to determine the mass of precipitate (PbIâ‚‚) that would be formed
From the balanced chemical equation,
1 mole of Pb(ClO₃)₂ reacts with 2 moles of NaI to produce 1 mole of PbI₂
First, we will determine the number of moles of NaI present
From the given information
Volume of NaI present = 0.400 L
Concentration of the NaI = 0.130 M
Using the formula
Number of moles = Concentration × Volume
Number of moles of NaI present = 0.130 × 0.400
∴ Number of moles of NaI present = 0.052 mole
Since,
1 mole of Pb(ClO₃)₂ reacts with 2 moles of NaI to produce 1 mole of PbI₂
Then,
The 1.50L concentrated Pb(ClO₃)₂ will react with 0.052 mole of NaI to produce [tex]\frac{0.052}{2}[/tex] mole of PbI₂
[tex]\frac{0.052}{2} = 0.026[/tex]
∴ Number of mole of PbI₂ formed is 0.026 mole
Now, for the mass of PbIâ‚‚ formed
Using the formula
Mass = Number of moles × Molar mass
Molar mass of PbIâ‚‚ = 461.009 g/mol
∴ Mass of PbI₂ formed = 0.026 × 461.009
Mass of PbIâ‚‚ formed = 11.986 g
Hence, the mass of precipitate that would form is 11.986 g
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