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

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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%