# How many grams of H2O2 were in the original sample?

Hydrogen peroxide (H2O2) decomposes in the presence of a catalyst to form water and oxygen. The catalyst was added to 5.60 mL of a hydrogen peroxide solution at 20.0 ∘C , and 49.5 mL of gas was collected over water at a total pressure of 764.4 mmHg . Look up the vapor pressure of water under these conditions in the table.

• WVP = 17.5 mm Hg

Gas pressure = 764.4 – 17.5 = 746.9 mm Hg

The temperature of the hydrogen peroxide solution must be converted from ˚C to ˚K by adding 273.

T = 20 + 273 = 293

Let the volume be converted to liters.

V = 0.0495 liter

Let’s use the following equation to equation to determine the volume of oxygen at standard temperature and pressure.

P1 V1 ÷ T1 = P2 V2 ÷ T2

746.9 0.0495 ÷ 293 = 760 V2 ÷ 273

V2 = 10,093.213315 ÷ 222,680

The volume is approximately 0.0453 liter. The volume of one mole of gas at standard temperature and pressure is 22.4 liters.

n = 10,093.213315 ÷ 222,680) ÷ 22.4

This is approximately 0.00202 mole of oxygen.

H2O2 → H2O + O2

To balance this equation, make 2 H2O2 and 2 H2O.

2 H2O2 → 2 H2O + O2

According to the coefficients in the balanced equation, two moles of hydrogen peroxide will produce two moles of water and one mole of oxygen. This means the number of moles hydrogen peroxide is twice the number of moles of oxygen.

n = (10,093.213315 ÷ 222,680) ÷ 11.2

This is approximately 0.00404 mole of hydrogen peroxide.

The mass of one mole of hydrogen peroxide is 34 grams.

This is approximately 0.138 grams. This is the only way that I know how to solve this type of problem. I hope this is helpful for you.

• Grams of H2O2.....

2H2O2(l) --> 2H2O(l) + O2(g)

...?g .............................. 49.5 mL, 20C, 764.4 Torr

1. Find the moles of O2 ............ use Dalton's law and the ideal gas equation

2. Double that to get moles of H2O2 ....... use the unit factor method for steps 2 and 3

3. Calculate mass of H2O2.

Vapor pressure of water at 20.0C is 17.5 Torr. Subtract to get the pressure of "dry" O2.

PV = nRT ................ ideal gas equation

n = PV / (RT) .......... solve for n (moles)

n = 746.9 Torr x 0.0495L / 62.36 LTorr/molK / 293.15K

n = 0.00202 mol O2

0.00202 mol O2 x (2 mol H2O2 / 1 mol O2) x (34.0g H2O2 / 1 mol H2O2) = 0.138g H2O2