The Fe2 (55.845 g/mol) content of a 2.252 g steel sample dissolved in 50.00 mL w
ID: 854942 • Letter: T
Question
The Fe2 (55.845 g/mol) content of a 2.252 g steel sample dissolved in 50.00 mL was determined by tiration with a standardized 0.100 M potassium permanganate (KMnO4, 158.034 g/mol) solution. The titration required 30.37 mL to reach the end point. What is the concentration of iron in the steel sample?
The Fe2 (55.845 g/mol) content of a 2.252 g steel sample dissolved in 50.00 mL was determined by tiration with a standardized 0.100 M potassium permanganate (KMnO4, 158.034 g/mol) solution. The titration required 30.37 mL to reach the end point. What is the concentration of iron in the steel sample?Explanation / Answer
No of moles of KMnO4 present in 30.37 ml of 0.100 M KMnO4 = 30.37*0.1*10-3 =3.037 milli moles
as each mole of KMnO4reacts with 5 moles of Fe2+ ,so no of moles of Fe2+ consumed =5*3.037 =15.185 milli moles
so weight of Fe2+ in the sample =55.845*15.185*10-3 =0.8480 gm
so concentration of iron in steel sample = (0.840/2.252)*100 =37.3 %
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.