the active ingredient in Antabuse, a drug used for the treatment of chronic alco

Question

the active ingredient in Antabuse, a drug used for the treatment of chronic alcoholism, is tetraethylthiuram disulfide, (296.54 g/mole). the sulfur in a 0.43210g sample of an Antabuse preparation was oxidized to SO2, which was absorbed in H2O2 to give H2SO4. The acid was titrated with 22.13 mL of 0.03736 M base. Calculate the percentage of active ingredient in the preparation. please show complete solutions with answer. don't give any sites.

Explanation / Answer

moles of base = 22.13 x0.03736/1000 = 0.0008267, moles of acid = 0.0008267/2 = 0.0004133 , since 1 H2So4 gives 2H+ , SO2+ H2O2 ---> H2SO4 , so SO2 moles = 0.0004133 , each tetraethylthiuram contains 4 S, so oxidation of 1 mol of compound gives 4 mol SO2, hence for 0.0004133 mol SO2 compound mol = 0.0004133/4 = 0.0001033, mass of compound = moles x mol mas = 0.0001033 x296.54 = 0.03064 gm , % of active ingrediant = (0.03064/0.4321) x100 = 7.09 %

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