What mass of hydrogen peroxide must be used to produce 1.00 L of oxygen gas at 25.0°C and 1.00 ATM?

2.79g

Explanation:

First let us obtain the number of mole of O2 produced at 1atm and 25°C. This is illustrated below:

V (volume) = 1L

P (pressure) = 1 atm

T (temperature) = 25°C = 25 + 273 = 298K

R (gas constant) = 0.082atm. L/Kmol

n (number of mole) = ?

Using the ideal gas equation PV = nRT, the number of mole can be obtained as follow:

PV = nRT

n = PV / RT

n = 1x1 / (0.082x298)

n = 0.041mole

Therefore the number of mole of O2 produced is 0.041mole.

Now let us generate a balanced equation for the decomposition of Hydrogen peroxide (H2O2) to produce oxygen (O2). This is shown below:

2H2O2 - > 2H2O + O2

From the equation,

2 moles of hydrogen peroxide (H2O2) produced 1 mole of oxygen (O2).

Therefore, Xmol of Hydrogen peroxide (H2O2) will produce 0.041mole of oxygen (O2) i. e

Xmol of Hydrogen peroxide (H2O2) = 0.041 x 2 = 0.082 mole.

Now let us calculate the mass of H2O2 in 0.082 mole of H2O2. This is shown below:

Molar Mass of H2O2 = (2x1) + (16x2) = 2 + 32 = 34g/mol

Number of mole of H2O2 = 0.082 mole

Mass of H2O2 = ?

Mass = number of mole x molar Mass

Mass of H2O2 = 0.082 x 34

Mass of H2O2 = 2.79g

Therefore, 2.79g of Hydrogen peroxide (H2O2) is used in the reaction