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Chloride anions in solution will combine with the silver cations to produce bright white silver chloride precipitate. Suppose an EPA chemist tests a 250.mL sample of groundwater known to be contaminated with iron(III) chloride, which would react with silver nitrate solution like this:FeCl3(aq) + 3AgNO3(aq) ⟶ 3AgCl(s) + FeNO3(aq)The chemist adds 82.0 M silver nitrate solution to the sample until silver chloride stops forming. He then washes, dries, and weighs the precipitate. He finds he has collected 2.5mg of silver chloride.Calculate the concentration of iron(III) chloride contaminant in the original groundwater sample.

Respuesta :

Answer:

[tex]2.326\times 10^{-5} M[/tex] the concentration of iron(III) chloride contaminant in the original groundwater sample.

Explanation:

[tex]FeCl_3(aq) + 3AgNO_3(aq)\rightarrow 3AgCl(s) + Fe(NO_3)_3(aq)[/tex]

Mass of silver chloride = 2.5 mg = 0.0025 g

1 mg = 0.001 g

Moles of silver chloride = [tex]\frac{0.0025 g}{143.32 g/mol}=1.744\times 10^{-5} mol[/tex]

According to reaction, 3 moles of silver chloride are obtained from 1 mole of ferric chloride, then [tex]1.744\times 10^{-5} mol[/tex] od silver chloride will be obtained from ;

[tex]\frac{1}{3}\times 1.744\times 10^{-5} mol=5.814\times 10^{-6} mol[/tex] of ferric chloride

Volume of the ground water sample = 250 mL= 0.250 L

1 mL = 0.001 L

Concentration of iron(III) chloride =

[tex][FeCl_3]=\frac{5.814\times 10^{-6} mol}{0.250 L}=2.326\times 10^{-5} M[/tex]

[tex]2.326\times 10^{-5} M[/tex] the concentration of iron(III) chloride contaminant in the original groundwater sample.