Answer:
There is 11.2 L of O2 produced
Explanation:
<u>Step 1:</u> Data given
Molar mass of H2O2 = 34.01 g/mol
Molar mass of O2 = 32 g/mol
Mass of H2O2 = 34.0 grams
<u>Step 2:</u> The balanced reaction
2H2O2 → 2H2O + O2
<u>Step 3</u>: Calculate number of moles H2O2
Moles H2O2 = Mass H2O2 / Molar mass H2O2
Moles H2O2 = 34 grams / 34.01 g/mol
Moles H2O2 = 1 mol
<u>Step 4: </u>Calculate the number of moles at the equilibrium
Initial number of moles H2O2 = 1 mol
Initial number of mol H2O and O2 = 0 mol
Since the mol ratio H2O2 H2O O2 is 2:2:1
There will react x mol of H2O2; X mol of H2O and 0.5 x mol of O2
At the equilibrium, H2O2 has (1-x) mol. Since it's decomposed into H2O and O2, we know x =1 mol
At the equilibrium, there is 1 mol H2O and 0.5 mol O2
Under STP, 1 mol of gas has a volume of 22.4 L
This means 0.5 moles of O2 has a volume of 22.4*0.5 = 11.2 L
There is 11.2 L of O2 produced
