Question #26490

1 Answer
Jun 17, 2017

Answer:

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#