Answer to Question #95758 in General Chemistry for Hassan Fallous

Question

The 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, 0.26 L of O2 gas is collected over water at a total pressure of 0.91 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? Remember, hydrogen peroxide is a liquid, not a gas.

Answer 1

0.91 atm - 0.0278 atm = 0.8822 atm (pressure of the O2)

PV = nRT

n = PV/RT = (0.8822 atm * 0.26 L) / (0.0821 Latm/Kmol * 296 K) = 0.229372 / 24.3016 = 0.00944 moles O2

0.00944 moles O2 x 2 moles H2O2/mole O2 x 34 g H2O2/mole = 0.64192 g H2O2 

0.64192 g H2O2