Hydrogen Peroxide, H2O2, in the presence of a catalyst decomposes into water and oxygen gas. How many L of O2 at STP are produced from the decomposition of 34.0g of H2O2? A.) 1.00B.) 5.60C.) 11.2D.) 22.4

Question

Chemistry

Jaylynn Powers

1 February, 13:29

Hydrogen Peroxide, H2O2, in the presence of a catalyst decomposes into water and oxygen gas. How many L of O2 at STP are produced from the decomposition of 34.0g of H2O2? A.) 1.00B.) 5.60C.) 11.2D.) 22.4

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Answers (1)

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Janet Fletcher · Answer 1

There is 11.2 L of O2 produced

Explanation:

Step 1: Data given

Molar mass of H2O2 = 34.01 g/mol

Molar mass of O2 = 32 g/mol

Mass of H2O2 = 34.0 grams

Step 2: The balanced reaction

2H2O2 → 2H2O + O2

Step 3: Calculate number of moles H2O2

Moles H2O2 = Mass H2O2 / Molar mass H2O2

Moles H2O2 = 34 grams / 34.01 g/mol

Moles H2O2 = 1 mol

Step 4: 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