How many liters of oxygen are released from the decomposition of 36.5 grams of hydrogen peroxide at STP?

12 liters of oxygen are released from the decomposition of the peroxide.

Explanation:

The hydrogen peroxide is decomposed by this reaction:

2 H₂O₂ (l) → 2 H₂O (l) + O₂ (g)

Molar mass H₂O₂ = 34 g/m

Moles of peroxide: 36.5 g / 34 g/m = 1.07 moles

Ratio is 2:1, so 2 moles of peroxide will be decomposed in 1 mol of oxygen (exactly the half)

1.07 moles of peroxide will be decomposed in the half of moles, 0.54 moles

Let's apply the Ideal Gases law Equation to solve the volume

1 atm. V = 0.54 mol. 0.082.273K

V = (0.54 mol. 0.082.273K) / 1 atm → 12L