Answer:
5.471% As₂O₃ in the sample.
Explanation:
<em>...the reaction is: Ag+ + SCN- => AgSCN(s) Calculate the percent As2O3 in the sample. (F.W. As2O3 = 197.84 g/mol).</em>
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First, with the amount of KSCN we can find the moles of Ag in the filtrates. As we know the amount of Ag added we can know the precipitate of Ag and the moles of AsO₄ = 1/2 moles of As₂O₃ in the sample:
<em>Moles KSCN = Moles Ag⁺ in the filtrate:</em>
0.01127L * (0.100mol / L)= 0.001127moles Ag⁺
<em>Total moles Ag⁺:</em>
0.0400L * (0.0781mol/L) = 0.0031564 moles Ag⁺
<em>Moles of Ag⁺ in the precipitate:</em>
0.0031564 - 0.001127 = 0.0020294 moles Ag⁺
<em>Moles AsO₄ = Moles As:</em>
0.0020294 moles Ag⁺ * (1mol As / 3 moles Ag⁺) = 6.765x10⁻⁴ moles AsO₄
<em>Moles As₂O₃:</em>
6.765x10⁻⁴ moles AsO₄ * (1 mol As₂O₃ / 2 mol AsO₄) =
3.382x10⁻⁴ moles As₂O₃
<em>Mass As₂O₃:</em>
3.382x10⁻⁴ moles As₂O₃ * (197.84g/mol) = 0.0669g As₂O₃
Percent is:
0.0669g As₂O₃ / 1.223g sample * 100 =
<h3>5.471% As₂O₃ in the sample</h3>
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