Precipitation (help)?
2016-06-15 2:21 pm
Assume that 100 ml of 2.00 M NaCl solution is mixed with 100 ml of 0, 0200 M AgNO3 solution. Determine the number of grams of AgCl which precipitates from solution.??
回答 (2)
2016-06-15 2:53 pm
✔ 最佳答案
NaCl(aq) + AgNO₃(aq) → AgCl(s) + NaNO₃(aq)Mole ratio NaCl : AgNO₃ = 1 : 1
No. of moles of NaCl = (2.00 mol/L) × (100/1000 L) = 0.200 mol
No. of moles of AgNO₃ = (0.0200 mol/L) × (100/1000 L) = 0.00200 mol < 0.200 mol
AgNO₃ is the limiting reactant.
According to the equation, mole ratio AgNO₃ : AgCl = 1 : 1
No. of moles of AgNO₃ reacted = 0.00200 mol
No. of moles of AgCl formed = 0.00200 mol
Molar mass of AgCl = (107.9 + 35.5) g/mol = 143.4 g/mol
Mass of AgCl formed = (143.4 g/mol) × (0.00200 mol) = 0.287 g
2016-06-15 3:15 pm
Ksp AgCl = 1.8×10–10
AgCl ---> Ag^+ + Cl^-
Ksp = [Ag+][Cl-]
1.8x10^-10 = [x][x] = x^2
x = 1.34x10^-5 M = solubility of the AgCl
Since 0.0020 moles AgCl were formed in a total volume of 200 ml, the [AgCl] = 0.01 M
1.34x10^-5 mol/L x 0.2 L = 2.68x10^-4 moles dissolved
0.0020 moles formed
0.0020 - 0.000268 moles = 0.001732 moles precipitated
0.001732 moles x 143.32 g/mol = 0.248 grams precipitated
AgCl ---> Ag^+ + Cl^-
Ksp = [Ag+][Cl-]
1.8x10^-10 = [x][x] = x^2
x = 1.34x10^-5 M = solubility of the AgCl
Since 0.0020 moles AgCl were formed in a total volume of 200 ml, the [AgCl] = 0.01 M
1.34x10^-5 mol/L x 0.2 L = 2.68x10^-4 moles dissolved
0.0020 moles formed
0.0020 - 0.000268 moles = 0.001732 moles precipitated
0.001732 moles x 143.32 g/mol = 0.248 grams precipitated
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