Question

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?

Answer 1

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:color(white)(m) 34.01`
`color(white)(mmm)"2H"_2"O"_2 → 2"H"_2"O" + "O"_2`

Step 2. Convert grams of `"H"_2"O"_2` to moles of `"H"_2"O"_2`

`"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 `"HCl"` to moles of `"O"_2`

`"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 `"O"_2`

For this calculation, we can use Ideal Gas Law:

`color(blue)(barul|stackrel(" ")( pV = nRT)|)`

WE can rearrange this formula to give

`V = (nRT)/p`

`n = "0.015 mol"`
`"R" = "0.082 06 L·atm·K"^"-1""mol"^"-1"`
`T = "(28 + 273.15) K = 301.15 K"`
`p = "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 `"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 `"H"_2"O"_2`.

However, I carried the calculation to two significant figures.

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