A 2.121 gram sample containing an unknown amount of arsenic trichloride and the

Question

A 2.121 gram sample containing an unknown amount of arsenic trichloride and the rest inerts was dissolved into a NaHCO3 and HCl aqueous solution. To this solution was added 1.740 grams of KI and 50.00 mL of a 0.00900 M KIO3 solution. The excess I3– was titrated with 50.00 mL of a 0.02000 M Na2S2O3 solution. What was the mass percent of arsenic trichloride in the original sample?A 2.121 gram sample containing an unknown amount of arsenic trichloride and the rest inerts was dissolved into a NaHCO3 and HCl aqueous solution. To this solution was added 1.740 grams of KI and 50.00 mL of a 0.00900 M KIO3 solution. The excess I3– was titrated with 50.00 mL of a 0.02000 M Na2S2O3 solution. What was the mass percent of arsenic trichloride in the original sample?

Explanation / Answer

moles of KIO3 = 0.009 M x 0.05 L = 4.5 x 10^-4 mols

1 mole of IO3- reacts to form 3 moles of I2

So, initial moles of I2 = 3 x 4.5 x 10^-3 = 1.35 x 10^-3 mols

moles of Na2S2O3 used = 0.02 M x 0.05 L = 1 x 10^-3 mols

2 moles of Na2S2O3 reacts with 1 mole of I2

so, moles of excess I2 = 1 x 10^-3/2 = 0.5 x 10^-3 moles

Now, moles of I2 reacted with AsCl3 = 1.35 x 10^-3 - 0.5 x 10^-3 = 8.5 x 10^-4 mols

1 mole of AsCl3 reacts with 1 mole of I2

moles of AsCl3 = 8.5 x 10^-4 mols

mass of AsCl3 reacted = 8.5 x 10^-4 x 181.28 = 0.154 g

mass percent of AsCl3 = (0.154/2.121) x 100 = 7.26%

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