Respuesta :
Answer:
D. The final concentration of NO3– is 0.821 M.
Explanation:
Considering:
[tex]Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}[/tex]
Or,
[tex]Moles =Molarity \times {Volume\ of\ the\ solution}[/tex]
Given :
For potassium iodide :
Molarity = 1.60 M
Volume = 300.0 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 300.0×10⁻³ L
Thus, moles of potassium iodide :
[tex]Moles=1.60 \times {300.0\times 10^{-3}}\ moles[/tex]
Moles of potassium iodide = 0.48 moles
For lead(II) nitrate :
Molarity = 1.20 M
Volume = 285 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 285×10⁻³ L
Thus, moles of lead(II) nitrate :
[tex]Moles=1.20\times {285\times 10^{-3}}\ moles[/tex]
Moles of lead(II) nitrate = 0.342 moles
According to the given reaction:
[tex]2KI_{(aq)}+Pb(NO_3)_2_{(aq)}\rightarrow PbI_2_{(s)}+2KNO_3_{(aq)}[/tex]
2 moles of potassium iodide react with 1 mole of lead(II) nitrate
1 mole of potassium iodide react with 1/2 mole of lead(II) nitrate
0.48 moles potassium iodide react with 0.48/2 mole of lead(II) nitrate
Moles of lead(II) nitrate = 0.24 moles
Available moles of lead(II) nitrate = 0.342 moles
Limiting reagent is the one which is present in small amount. Thus, potassium iodide is limiting reagent.
Also, consumed lead(II) nitrate = 0.24 moles (lead ions precipitate with iodide ions)
Left over moles = 0.342 - 0.24 moles = 0.102 moles
Total volume = 300 + 285 mL = 585 mL = 0.585 L
So, Concentration = 0.102/0.585 M = 1.174 M
Statement A is correct.
The formation of the product is governed by the limiting reagent. So,
2 moles of potassium iodide gives 1 mole of lead(II) iodide
1 mole of potassium iodide gives 1/2 mole of lead(II) iodide
0.48 mole of potassium iodide gives 0.48/2 mole of lead(II) iodide
Mole of lead(II) iodide = 0.24 moles
Molar mass of lead(II) iodide = 461.01 g/mol
Mass of lead(II) chloride = Moles × Molar mass = 0.24 × 461.01 g = 111 g
Statement B is correct.
Potassium iodide is the limiting reagent. So all the potassium ion is with potassium nitrate . Thus,
2 moles of Potassium iodide on reaction forms 2 moles of potassium ion
0.48 moles of Potassium iodide on reaction forms 0.48 moles of potassium ion
Total volume = 300 + 285 mL = 585 mL = 0.585 L
So, Concentration = 0.48/0.585 M = 0.821 M
Statement C is correct.
Nitrate ions are furnished by lead(II) nitrate . So,
1 mole of lead(II) nitrate produces 2 moles of nitrate ions
0.342 mole of lead(II) nitrate produces 2*0.342 moles of nitrate ions
Moles of nitrate ions = 0.684 moles
So, Concentration = 0.684/0.585 M = 1.169 M
Statement D is incorrect.
Statements A, B and C are correct but statement D is incorrect.
The equation of the reaction is as follows;
Pb(NO3)2(aq) + 2KI(aq) = PbI2(s) + 2KNO3(aq)
Number of moles of Pb(NO3)2 = 1.20 M × 285.0/1000 L
= 0.342 moles
Number of moles of KI = 1.60 M × 300.0//1000 L
= 0.48 moles
Since 1 mole of Pb(NO3)2 reacts with 2 moles of KI
0.342 moles of Pb(NO3)2 reacts with 0.342 moles × 2 moles/1 mole
= 0.684 moles hence KI is the limiting reactant
2 moles of KI yields 1 mole of PbI2
0.48 moles of KI yields 0.48 moles × 1 mole /2 moles = 0.24 moles
Mass of PbI2 produced = 0.24 moles × 461 g/mol = 111 g of lead(II) iodide.
Hence, it is true that you form 111 g of lead(II) iodide.
Amount of lead II ions left over = 0.342 moles - 0.24 moles = 0.102 moles
Total volume = 285.0 mL + 300.0 mL = 585 mL or 0.585 L
Final concentration of lead II ions = 0.102 moles/ 0.585 L = 0.174 M
The statement is correct that the final concentration of Pb2+ ions is 0.174 M.
Since 2 moles of KI produces 2 moles of KNO3
0.48 moles of KI produces 0.48 moles of KNO3
Final concentration of potassium ion = 0.48 moles/0.585 L = 0.821 M
It is true that the final concentration of K+ is 0.821 M.
1 mole of lead(II) nitrate produces 2 moles of nitrate ions
0.342 mole of lead(II) nitrate produces 0.684 moles of lead(II) nitrate
Final concentration of nitrate ion = 0.684 moles/0.585 L = 1.169 M
The statement that the final concentration of NO3– is 0.821 M is incorrect.
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