a) Balanced equation
#5×["H"_2"O"_2 → "O"_2 + "2H"^"+" + 2"e"^"-"]#
#2×["MnO"_4^"-" + "8H"^"+" + 5"e"^"-" → "Mn"^"2+" + 4"H"_2"O"]#
#bb(5"H"_2"O"_2 + "2MnO"_4^"-" + "6H"^"+" → "5O"_2 + "2Mn"^"2+" + 8"H"_2"O")#
b) Moles of #"MnO"_4^"-"#
#"Moles of MnO"_4^"-" = "0.024 75" color(red)(cancel(color(black)("L MnO"_4^"-"))) × ("0.0215 mol MnO"_4^"-")/(1 color(red)(cancel(color(black)("L MnO"_4^"-")))) = bb(5.321 × 10^"-4"color(white)(l) "mol MnO"_4^"-")#
c) Moles of #"H"_2"O"_2#
#"Moles of H"_2"O"_2 =5.321 × 10^"-4" color(red)(cancel(color(black)("mol MnO"_4^"-"))) × ("5 mol H"_2"O"_2)/(2 color(red)(cancel(color(black)("mol MnO"_4^"-")))) = bb(1.330 × 10^"-3" color(white)(l)"mol H"_2"O"_2)#
d) Molarity of #"H"_2"O"_2# in Sol A
Sol A was diluted from 25 mL to 250 mL, and you titrated a 25 mL aliquot of the dilute solution.
#"Moles of H"_2"O"_2 = 1.330 × 10^"-3"color(white)(l) "mol H"_2"O"_2 × (250 color(red)(cancel(color(black)("mL"))))/(25 color(red)(cancel(color(black)("mL")))) = "0.0133 mol H"_2"O"_2#
#"Molarity" = "0.0133 mol"/"0.250 L" = bb"0.0532 mol/L"#
e) Volume of #"O"_2#
#2"H"_2"O"_2 → 2"H"_2"O" + "O"_2#
#"Moles of H"_2"O"_2 = 1 color(red)(cancel(color(black)("L H"_2"O"_2))) × ("0.0532 mol H"_2"O"_2)/(1 color(red)(cancel(color(black)("L H"_2"O"_2)))) = "0.0532 mol H"_2"O"_2#
#"Moles of O"_2 = 0.0532 "mol H"_2"O"_2 × "1 mol O"_2/(2 "mol H"_2"O"_2) = "0.0266 mol O"_2#
From the Ideal Gas Law,
#color(blue)(bar(ul(|color(white)(a/a)pV = nRTcolor(white)(a/a)|)))" "#
so.
#V = (nRT)/p#
STP is defined as 1 bar and 0 °C.
∴ #V = (0.0266 color(red)(cancel(color(black)("mol"))) × "0.083 14" color(red)(cancel(color(black)("bar")))·"L"·color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 273.15 color(red)(cancel(color(black)("K"))))/(1 color(red)(cancel(color(black)("bar")))) = "0.604 L" = bb"604 mL"#
