Question

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?

Answer 1

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

Step-by-step 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:

`n_{H_(2)O_(2)} = (m)/(M) = (17.0 g)/(34 g/mol) = 0.5 moles`

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

`2n_{H_(2)O_(2)} = 2n_{H_(2)O} \rightarrow n_{H_(2)O} = 0.5 moles`

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

`V = (nRT)/(P) = (0.5 moles*0.085 L*atm/K*mol*273 K)/(1 atm) = 11.60 L`

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

I hope it helps you!