very confusing chem question

Question

he following reaction illustrates the decomposition of hydrogen peroxide (H2O2) to form water and oxygen gas:

2 H2O2 (l) → 2 H2O (l) + O2 (g)

When this reaction is carried out, 0.50 L of O2 gas is collected over water at a total pressure of 1.05 atm. The temperature is 23°C. The vapour pressure of water at this temperature is 0.0278 atm.

What was the mass (in g) of the hydrogen peroxide that decomposed?

Answer 1

When a gas is collected over water, one must subtract the vapor pressure of water from the total pressure in order to find the pressure of the gas itself.  Thus,  1.05 atm - 0.0278 atm = 1.0222 atm = pressure of the O₂ gas.

 

PV = nRT and solve for n to determine moles of O₂ collected.

n = PV/RT = (1.0222 atm)(0.5 L)/(0.0821 Latm/Kmol)(296 K)  note the use of Kelvin for temperature.

n = 0.02103 moles O₂

mass of O₂ = 0.02103 moles x 32 g/mole = 0.67 grams (to 2 significant figures)