# Question fc809

Dec 10, 2016

The volume of oxygen is 576 L.

#### Explanation:

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

(a) Mass of ${\text{H"_2"O}}_{2}$

${\text{Mass of H"_2"O"_2 = 1270 color(red)(cancel(color(black)("mL H"_2"O"_2))) × ("1.00 g H"_2"O"_2)/(1 color(red)(cancel(color(black)("mL H"_2"O"_2)))) = "1270 g H"_2"O}}_{2}$

(b) Moles of ${\text{H"_2"O}}_{2}$

1270 color(red)(cancel(color(black)("g H"_2"O"_2))) × ("1 mol H"_2"O"_2)/(34.01 color(red)(cancel(color(black)("g H"_2"O"_2)))) = "37.34 mol H"_2"O"_2 

(b) Moles of ${\text{O}}_{2}$

37.34 color(red)(cancel(color(black)("mol H"_2"O"_2))) × ("1 mol O"_2)/(2 color(red)(cancel(color(black)("mol H"_2"O"_2)))) = "18.67 mol O"_2

(c) Volume of ${\text{O}}_{2}$

The Ideal Gas Law is

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

We can rearrange this to give

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

$n = {\text{18.67 mol O}}_{2}$
$R = \text{0.082 06 L·atm·K"^"-1""mol"^"-1}$
$T = \text{(78 + 273.15) K" = "351.15 K}$
$P = \text{0.934 atm}$

V = (18.67 color(red)(cancel(color(black)("mol"))) × "0.082 06 L"·color(red)(cancel(color(black)("atm·K"^"-1""mol"^"-1"))) × 351.15 color(red)(cancel(color(black)( "K"))))/(0.934 color(red)(cancel(color(black)("atm")))) = "576 L"#

The volume of ${\text{O}}_{2}$ produced is $\text{576 L}$.