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?

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1 Answer
Sep 26, 2017

Answer:

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.