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A 0.477 mol sample of O_2 gas has a volume of 11.3 L at a certain temperature and pressure. If all this O_2 were converted to ozone (O_3) at the same temperature and pressure, what is the ozone volume (in liters)? 3 O_2(g) → 2 O_3(g)

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whitneytr12

Answer:

The volume of ozone produced is 7.53 L.

Explanation:

The reaction is:

3O₂(g)  →  2O₃(g)   (1)      

0.477 mol   V=?

11.3 L    

From the reaction (1) we have that 3 moles of O₂ produce 2 moles of O₃ so the volume of the ozone produced can be calculated as follows:

[tex] V_{O_{3}} = V_{O_{2}}*\frac{n_{O_{3}}}{n_{O_{2}}} = 11.3 L*\frac{2}{3} = 7.53 L [/tex]

Therefore, the volume of ozone produced is 7.53 L.

I hope it helps you!

The ozone volume (in liters) is 7.53 L.

  • The calculation is as follows:

The volume of the ozone in liters should be

[tex]= 11.3 \times 2\div 3[/tex]

= 7.53L

we have that 3 moles of O₂ produce 2 moles of O₃

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