Answer:
2.4 × 10^{-4} M Fe(II) or 1.3 × 10^{-2} mg Fe(II)/mL
Explanation:
According with the stoichiometry of the reaction:
[tex]6.9 mg AgCl (\frac{1mmol AgCl}{143.32 mg AgCl})(\frac{1 mmol FeCl_{2} }{1mmol AgCl})(\frac{1mmol Fe}{1mmol FeCl_{2} })= 0.048144 mmol Fe(II)\\[/tex]
Now, we divide by the volume of the aliquot
[tex]\frac{0.048144 mmol Fe(II)}{200 mL sample}= 2.4x10^{-4} M Fe(II)[/tex]
Answer:
The concentration of iron(II) chloride contaminant in the original groundwater sample is [tex]1.2\times 10^{-4} mol/L[/tex].
Explanation:
[tex]FeCl_2 (aq) + 2AgNO_3 (aq)\rightarrow 2AgCl (s) + Fe(NO_3)_2 (aq)[/tex]
Amount of silver chloride recovered = 6.9 mg = 0.0069 g[/tex]
Moles of silver chloride = [tex]\frac{0.0069 g}{143.5 g/mol}=4.8252\times 10^{-5} mol[/tex]
According to reaction. 2 moles of silver chloride are obtained from 1 mole of iron(II) chloride.
Then [tex]4.8252\times 10^{-5}[/tex] moles of silver chloride will be obtained from:
[tex]\frac{1}{2}\times 4.8252\times 10^{-5} mol=2.4126\times 10^{-5} mol[/tex]
Moles of iron(II) chloride = [tex]2.4126\times 10^{-5} mol[/tex]
Volume of a sample of ground water = 200 mL = 0.2 L
[tex]Concentration=\frac{Moles}{Volume(L)}[/tex]
[tex][FeCl_2]=\frac{2.4126\times 10^{-5} mol}{0.2 L}=0.00012063 mol/L[/tex]
[tex][FeCl_2]=1.2063\times 10^{-4} mol/L\approx 1.2\times 10^{-4} mol/L[/tex]
The concentration of iron(II) chloride contaminant in the original groundwater sample is [tex]1.2\times 10^{-4} mol/L[/tex].
