Answer:
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:
[tex] n_{H_{2}O_{2}} = \frac{m}{M} = \frac{17.0 g}{34 g/mol} = 0.5 moles [/tex]
From the reaction (1) we have that 2 moles of H₂O₂ produce 2 moles of H₂O, so the number of moles of H₂O is:
[tex] 2n_{H_{2}O_{2}} = 2n_{H_{2}O} \rightarrow n_{H_{2}O} = 0.5 moles [/tex]
Now, we can find the volume of water vapor produced at STP, using Ideal gas law:
[tex] V = \frac{nRT}{P} = \frac{0.5 moles*0.085 L*atm/K*mol*273 K}{1 atm} = 11.60 L [/tex]
Therefore, the volume of water vapor produced at STP is 11.60 L.
I hope it helps you!
