Question

A chemical engineer determines the mass percent of iron in an ore sample by converting the fe to fe2+ in acid and then titrating the fe2+ with mno4−. a 1.3909−g sample was dissolved in acid and then titrated with 21.63 ml of 0.04462 m kmno4. the balanced equation is

Answer 1

Answer : The question is incomplete.

The complete question will be -

A chemical engineer determines the mass percent of iron in an ore sample by converting the Fe to
`Fe^(2+)` in acid and then titrating the
`Fe^(2+)` with Mn
`O^(4-)`.
A 1.3909 g sample was dissolved in acid and then titrated with 21.63 mL of 0.03709 M KMn
`O^(4-)`. The balanced equation is given below.
8
`H^(+)`(aq) + 5
`Fe^(2+)`(aq) + Mn
`O^(4-)`(aq) → 5
`Fe^(3+)`(aq) +
`Mn^(+2)` +(aq) + 4
`H_(2)O`(l)
Calculate the mass percent of iron in the ore.
______%

Calculation : Given :- weight of sample - 1.3909 g,
Moles of KMn
`O^(4-)` - 0.04462 m
Volume of KMn
`O^(4-)` - 21.63 mL

On solving we get,
(21.63 mL KMn
`O^(4-)` / 1.3909g sample) x (1L / 1000mL) x (0.04462mol KMn
`O^(4-)` / L KMn
`O^(4-)`) x (5 mol Fe2+ / 1 mol KMn
`O^(4-)`) x (55.845g
`Fe^(2+)` / mol
`Fe^(2+)`) x 100% = 19.37%

So the concentration of the ore is 19.37%