Achemical 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 : 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  in acid and then titrating the  with Mn.
A 1.3909 g sample was dissolved in acid and then titrated with 21.63 mL of 0.03709 M KMn. The balanced equation is given below. 
8 (aq) + 5 (aq) + Mn(aq) → 5 (aq) +  +(aq) + 4 (l) 
Calculate the mass percent of iron in the ore. 
______%

Calculation :  Given :- weight of sample - 1.3909 g,
Moles of KMn - 0.04462 m
Volume of KMn - 21.63 mL

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

So the concentration of the ore is 19.37% 

The answer to that would be magnesium