Question

The iron content of iron ore can be determined by titration with a standard solution. The iron ore is dissolved in , and all the iron is reduced to ions. This solution is then titrated with solution, producing and ions in acidic solution. If it required 16.37 mL of 0.0233 M to titrate a solution made from 0.3298 g of iron ore, what is the mass percent of iron in the iron ore?

Answer 1

Answer:

2.11 %

Explanation:

In acidic medium iron is in Fe ⁺³ oxidation state .

Equivalent weight = 56 / 3

= 18.33 gm

acid used in titration

= 16.37 mL of .0233 M

= 16.37 x .0233 mL of M soln

= .38 mL of M soln

.38 mL of M soln reacts with .3298 gm of ore

1000 mL of M soln = (.3298 / .38)  x 1000 gm of iron ore

867.89 gm

This must contain one gm equivalent of iron

or 18.33 gm

867.89 gm of ore contains 18.33 gm of iron

mass % of iron in the given ore

= (18.33 / 867.89)  x 100

= 2.11 %