# Question b9b56

Jun 18, 2017

Here's what I get.

#### Explanation:

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"]
$\boldsymbol{5 \text{H"_2"O"_2 + "2MnO"_4^"-" + "6H"^"+" → "5O"_2 + "2Mn"^"2+" + 8"H"_2"O}}$

b) Moles of $\text{MnO"_4^"-}$

"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 ${\text{H"_2"O}}_{2}$

"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 ${\text{H"_2"O}}_{2}$ in Sol A

Sol A was diluted from 25 mL to 250 mL, and you titrated a 25 mL aliquot of the dilute solution.

${\text{Moles of H"_2"O"_2 = 1.330 × 10^"-3"color(white)(l) "mol H"_2"O"_2 × (250 color(red)(cancel(color(black)("mL"))))/(25 color(red)(cancel(color(black)("mL")))) = "0.0133 mol H"_2"O}}_{2}$

$\text{Molarity" = "0.0133 mol"/"0.250 L" = bb"0.0532 mol/L}$

e) Volume of ${\text{O}}_{2}$

$2 {\text{H"_2"O"_2 → 2"H"_2"O" + "O}}_{2}$

${\text{Moles of H"_2"O"_2 = 1 color(red)(cancel(color(black)("L H"_2"O"_2))) × ("0.0532 mol H"_2"O"_2)/(1 color(red)(cancel(color(black)("L H"_2"O"_2)))) = "0.0532 mol H"_2"O}}_{2}$

${\text{Moles of O"_2 = 0.0532 "mol H"_2"O"_2 × "1 mol O"_2/(2 "mol H"_2"O"_2) = "0.0266 mol O}}_{2}$

From the Ideal Gas Law,

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} p V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

so.

$V = \frac{n R T}{p}$

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"#