# Calculate the volume of O2(g) produced at standard conditions of temperature and pressure.?

The catalytic decomposition of hydrogen peroxide can be expressed as:

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

A trial of this decomposition experiment produced the following data:

Volume of O2 produced at room conditions 240 ml

Barometric pressure 740 torr

Temperature of water 24oC

Temperature of O2 25oC

Vapor Pressure due to water at 24oC 22.4 torr

Calculate the volume of O2(g) produced at standard conditions of temperature and pressure. (enter your answer in liters)

Relevance

As you may know, O2 is collected over water, thus water vapor is saturated in the gas we collected.

The given information tells us:

Temperature of water 24oC, and vapor Pressure due to water at 24oC 22.4 torr

Volume of O2 produced at room conditions 240 ml

Barometric pressure 740 torr = water partial pressure plus O2 partial pressure <==> O2 pressure = 740 - 22.4 = 717.6 (torr)

Temperature of O2 25oC

You also know that the standard condition is 760 torr and 0 degree C. Therefore by using ideal gas law:

PV/T = pv/t

V = pvT/Pt = 717.6*240*273.15/ (760*298.15) = 208 (ml)

• This is fully in conformance with Avagadro. For any gases, as long as they are all at the same conditions of temperature and pressure, mols can be read as volumes. This is true under all equal conditions, not only at STP. Example:( this is purely theoretical, that the reaction goes to completion, which it does not.) N2+ 3H2 → 2NH3 You can read this as: 1mol of nitrogen will react with 3mol of hydrogen to produce 2 mol ammonia. Or you can equally correctly say: 1volume of nitrogen reacts with 3 volumes of nitrogen to produce 2 volumes of ammonia. Pressure and temperature are irrelevant, as long as all three gases are at the same conditions.