Answer:
[tex]V_{O2}= 1.5m^3[/tex]
Explanation:
From the question we are told that:
Volume of H_2O consumed [tex]V_{H2O}=3.0cm^3[/tex]
Balanced equation of Reaction is
[tex]2 F_2 + 2 H_2O -> 4 HF + O_2[/tex]
Therefore
O_2 produced in Volume is
[tex]V_{O2}= \frac{1}{2}*V_{H2O}[/tex]
[tex]V_{O2}= \frac{1}{2}*3.0m^3[/tex]
[tex]V_{O2}= 1.5m^3[/tex]
