Question

The Fe2+ (55.845 g/mol) content of a 2.264 g steel sample dissolved in 50.00 mL of an acidic solution was determined by tiration with a standardized 0.120 M potassium permanganate (KMnO4, 158.034 g/mol) solution. The titration required 44.82 mL to reach the end point. What is the concentration of iron in the steel sample? Express your answer as grams of Fe per gram of steel Mn2+5 Fe34 H20 + 5 Fe2+ MnO8 H concentration g Fe/g steel A 1.969 g sample containing an unknown amount of arsenic trichloride, with the rest being inerts, was dissolved into a NaHCO and HC aqueous solution. To this solution was added 1.520 g of KI and 50.00 mL of a 0.00934 M KIO solution Iwas titrated with 50.00 mL of a 0.02000 M Na,S,03 solution The excess V What is the mass percent of arsenic trichloride in the original sample? mass percent

Answer 1

Sol 1.

As Reaction :

MnO4- + 8H+ + 5Fe2+ <---> Mn2+ + 5Fe3+ + 4H2O

Moles of MnO4- ( KMnO4) = Conc. of KMnO4 × Volume of KMnO4 / 1000  

= 0.120 × 44.82 / 1000 = 0.0053784 mol

From reaction , 1 mole of MnO4- combines with 5 moles of Fe2+

So , 0.0053784 moles of MnO4- combines with

= 0.0053784 × 5 = 0.026892 moles of Fe2+

As Molar Mass of Fe2+ = 55.845 g/mol

So , Mass of iron (Fe2+ ) = 0.026892 × 55.845 = 1.5017 g

As Mass of steel sample = 2.264 g

So , Concentration of iron in the steel sample

= Mass of iron / Mass of steel sample

= 1.5017 / 2.264  

= 0.6633 g Fe / g steel