Question

Question #26490

Answer 1

1 mol of `"FeC"_2"O"_4` will reduce b) 3/5 mol of `"KMnO"_4`.

Explanation:

Write the balanced equation

`3×["MnO"_4^"-" + "8H"^"+" + 5"e"^"-" → "Mn"^"2+" + 4"H"_2"O"]`
`5×["FeC"_2"O"_4 → "Fe"^"3+" + "2CO"_2 + 3"e"^"-"]`
`stackrel(—————————————————————)("3MnO"_4^"-" + "5FeC"_2"O"_4 + "24H"^"+" → "3Mn"^"2+" + "5Fe"^"3+" + "10CO"_2 + 12"H"_2"O")`

`"3KMnO"_4 + "5FeC"_2"O"_4 + "24H"^"+" → "3Mn"^"2+" + "5Fe"^"3+" + "3K"^"+" + "10CO"_2 + 12"H"_2"O"`

Calculate the moles of `"KMnO"_4`

`"Moles of KMnO"_4 = 1 color(red)(cancel(color(black)("mol FeC"_2"O"_4))) × "3 mol KMnO"_4/(5 color(red)(cancel(color(black)("mol FeC"_2"O"_4)))) = 3/5 color(white)(l)"mol KMnO"_4`