Hydrogen peroxide (H2O2, 34 g/mol) decomposes into water vapor and oxygen gas. How many liters of water vapor are produced from the decomposition of 17.0 g of H2O2 at STP?

The volume of water vapor produced at STP is 11.60 L.

Explanation:

The decomposition reaction of H₂O₂ is:

2H₂O₂ → O₂ + 2H₂O     (1)

We have:

m(H₂O₂): mass = 17.0 g

M(H₂O₂): molar mass = 34 g/mol    

To find the volume of water vapor produced we need to find first the number of moles of H₂O₂ and H₂O:

From the reaction (1) we have that 2 moles of H₂O₂ produce 2 moles of H₂O, so the number of moles of H₂O is:

Now, we can find the volume of water vapor produced at STP, using Ideal gas law:

Therefore, the volume of water vapor produced at STP is 11.60 L.

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