How many liters of oxygen gas can be produced at stp from the decomposition of 0.250 l of 3.00 m h2o2 in the reaction according to the chemical equation shown below? 2h2o2 (l) → 2h2o (l) + o2 (g) ?

the balanced chemical equation for the decomposition of H₂O₂ is as follows

2H₂O₂ - --> 2H₂O + O₂

stoichiometry of H₂O₂ to O₂ is 2:1

the number of moles of H₂O₂ decomposed is - 0.250 L x 3.00 mol/L = 0.75 mol

according to stoichiometry the number of O₂ moles is half the number of H₂O₂ moles decomposed

number of moles of O₂ - 0.75 mol / 2 = 0.375 mol

apply the ideal gas law equation to find the volume

PV = nRT

where P - standard pressure - 10⁵ Pa

V - volume

n - number of moles 0.375 mol

R - universal gas constant - 8.314 Jmol⁻¹K⁻¹

T - standard temperature - 273 K

substituting the values in the equation

10⁵ Pa x V = 0.375 mol x 8.314 Jmol⁻¹K⁻¹ x 273 K

V = 8.5 L

volume of O₂ gas is 8.5 L