The sulfur in a 5.00 g sample of steel was evolved as H 2 S and then collected i

Question

The sulfur in a 5.00 g sample of steel was evolved as H2S and then collected in a solution of CdCl2 to produce the precipitate CdS. The CdS was then titrated with excess I2 and the remaining I2 was then back titrated with 4.82 ml of 0.0510 M sodium thiosulfate. If the concentration of the I2 is 0.0600 M and a total of 10.0 ml is added, calculate the %S in the steel. The relevant reactions are:

CdS + I2 ? S + Cd2+ + 2I-     Titration (excess I2)

I2 + 2S2O32- ? S2O62- + 2I     -Back Titration

Please show all steps and calculations...and correct amount of significant figures :)

Explanation / Answer

no. of millimoles of I2 used = 0.0600*10 - .0510*4.82 =0.35418 millimoles

moles of I2 used = moles of cds= moles of sulphur= .35418 millimoles

mass of sulphur = 32* .35418 = 11.33 mg

% s in steel= 11.33 * 100/(5*1000) =0.2266%

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