# How many liters of oxygen will be obtained from the decomposition of 1 g of H2O2. if the gas is collected at 28 degrees celsius and 0.867 atm?

Sep 26, 2017

You will obtain 0.42 L of oxygen.

#### Explanation:

There are four steps involved in this stoichiometry problem:

Step 1. Write the balanced chemical equation.

${M}_{r} : \textcolor{w h i t e}{m} 34.01$
$\textcolor{w h i t e}{m m m} {\text{2H"_2"O"_2 → 2"H"_2"O" + "O}}_{2}$

Step 2. Convert grams of ${\text{H"_2"O}}_{2}$ to moles of ${\text{H"_2"O}}_{2}$

${\text{Moles of H"_2"O"_2 = 1 color(red)(cancel(color(black)("g H"_2"O"_2))) × ("1 mol H"_2"O"_2)/(34.01 color(red)(cancel(color(black)("g H"_2"O"_2)))) = "0.029 mol H"_2"O}}_{2}$

Step 3. Convert moles of $\text{HCl}$ to moles of ${\text{O}}_{2}$

${\text{Moles of O"_2 = 0.029color(red)(cancel(color(black)("mol H"_2"O"_2))) × ("1 mol O"_2)/(2 color(red)(cancel(color(black)("mol H"_2"O"_2)))) = "0.015 mol O}}_{2}$

Step 4. Calculate the volume of ${\text{O}}_{2}$

For this calculation, we can use Ideal Gas Law:

$\textcolor{b l u e}{\overline{\underline{|}} \stackrel{\text{ }}{p V = n R T} |}$

WE can rearrange this formula to give

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

$n = \text{0.015 mol}$
$\text{R" = "0.082 06 L·atm·K"^"-1""mol"^"-1}$
$T = \text{(28 + 273.15) K = 301.15 K}$
$p = \text{0.867 atm}$

V= (0.015 color(red)(cancel(color(black)("mol"))) × "0.082 06 L"·color(red)(cancel(color(black)("atm·K"^"-1""mol"^"-1"))) × 301.15 color(red)(cancel(color(black)("K"))))/(0.867 color(red)(cancel(color(black)("atm")))) = "0.42 L"

The volume of ${\text{O}}_{2}$ formed is 0.42 L.

Note: The answer can have only one significant figure, because that is all you gave for the mass of ${\text{H"_2"O}}_{2}$.

However, I carried the calculation to two significant figures.

Here's a useful video on mass-volume conversions.