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