Respuesta :
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 [tex] Fe^{2+} [/tex] in acid and then titrating the [tex] Fe^{2+} [/tex] with Mn[tex] O^{4-} [/tex].
A 1.3909 g sample was dissolved in acid and then titrated with 21.63 mL of 0.03709 M KMn[tex] O^{4-} [/tex]. The balanced equation is given below.Â
8 [tex] H^{+} [/tex](aq) + 5 [tex] Fe^{2+} [/tex](aq) + Mn[tex] O^{4-} [/tex](aq) → 5 [tex] Fe^{3+} [/tex](aq) + [tex] Mn^{+2} [/tex] +(aq) + 4 [tex] H_{2}O [/tex](l)Â
Calculate the mass percent of iron in the ore.Â
______%
Calculation :Â Given :- weight of sample - 1.3909 g,
Moles of KMn[tex] O^{4-} [/tex] - 0.04462 m
Volume of KMn[tex] O^{4-} [/tex] - 21.63 mL
On solving we get,
(21.63 mL KMn[tex] O^{4-} [/tex]Â / 1.3909g sample) x (1L / 1000mL) x (0.04462mol KMn[tex] O^{4-} [/tex]Â / L KMn[tex] O^{4-} [/tex]) x (5 mol Fe2+ / 1 mol KMn[tex] O^{4-} [/tex]) x (55.845g [tex] Fe^{2+} [/tex]Â / mol [tex] Fe^{2+} [/tex]) x 100% = 19.37%Â
So the concentration of the ore is 19.37%Â
The complete question will be -
A chemical engineer determines the mass percent of iron in an ore sample by converting the Fe to [tex] Fe^{2+} [/tex] in acid and then titrating the [tex] Fe^{2+} [/tex] with Mn[tex] O^{4-} [/tex].
A 1.3909 g sample was dissolved in acid and then titrated with 21.63 mL of 0.03709 M KMn[tex] O^{4-} [/tex]. The balanced equation is given below.Â
8 [tex] H^{+} [/tex](aq) + 5 [tex] Fe^{2+} [/tex](aq) + Mn[tex] O^{4-} [/tex](aq) → 5 [tex] Fe^{3+} [/tex](aq) + [tex] Mn^{+2} [/tex] +(aq) + 4 [tex] H_{2}O [/tex](l)Â
Calculate the mass percent of iron in the ore.Â
______%
Calculation :Â Given :- weight of sample - 1.3909 g,
Moles of KMn[tex] O^{4-} [/tex] - 0.04462 m
Volume of KMn[tex] O^{4-} [/tex] - 21.63 mL
On solving we get,
(21.63 mL KMn[tex] O^{4-} [/tex]Â / 1.3909g sample) x (1L / 1000mL) x (0.04462mol KMn[tex] O^{4-} [/tex]Â / L KMn[tex] O^{4-} [/tex]) x (5 mol Fe2+ / 1 mol KMn[tex] O^{4-} [/tex]) x (55.845g [tex] Fe^{2+} [/tex]Â / mol [tex] Fe^{2+} [/tex]) x 100% = 19.37%Â
So the concentration of the ore is 19.37%Â
Mass percent is the percentage of the mass of the solute in the compound. The mass percent of iron in the ore sample is 19.37 %.
What is mass percent?
The mass percentage is the ratio of the mass of the solute to the mass of the compound multiplied by 100. Â The formula for the mass percent can be given as,
[tex]\rm mass \% = \dfrac{\text{molar mass of element}}{\text{molecular mass of compound}} \times 100[/tex]
Given,
Mass of the sample = 1.3909 gm
Moles of Potassium permanganate = 0.04462 mol
Volume of Potassium permanganate = 21.63 mL
The concentration of the iron in the ore is calculated as:
[tex]\begin{aligned} &= (\dfrac{21.63}{1.3909}) \times (\rm \dfrac{1L}{1000\;mL}) \times (0.04462) \times (\dfrac{5}{1}) \times (55.845) \times 100\% \\\\&= 19.37\% \end{aligned}[/tex]
Therefore, 19.37% is the mass percent of iron in the ore.
Learn more about mass percent here:
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