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.
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)
Calculate the mass percent of iron in the ore.
______%
Calculation : 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] / 1.3909g sample) x (1L / 1000mL) x (0.04462mol KMn[tex] O^{4-} [/tex] / L KMn[tex] O^{4-} [/tex]) x (5 mol Fe2+ / 1 mol KMn[tex] O^{4-} [/tex]) x (55.845g [tex] Fe^{2+} [/tex] / mol [tex] Fe^{2+} [/tex]) x 100% = 19.37%
So the concentration of the ore is 19.37%
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.
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)
Calculate the mass percent of iron in the ore.
______%
Calculation : 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] / 1.3909g sample) x (1L / 1000mL) x (0.04462mol KMn[tex] O^{4-} [/tex] / L KMn[tex] O^{4-} [/tex]) x (5 mol Fe2+ / 1 mol KMn[tex] O^{4-} [/tex]) x (55.845g [tex] Fe^{2+} [/tex] / mol [tex] Fe^{2+} [/tex]) x 100% = 19.37%
So the concentration of the ore is 19.37%
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. 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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