Chemistry - Mole Concept Question with Solution | TestHub

The maximum number of moles is calculated as:

$n_{max}=\frac{1120}{22400}=0.05$

Using the given values, we can set up the following equation:

$$\frac{x}{32}+\frac{1.76-x}{48}=0.05$$

Solving for $x$, we find $x=1.28$.

The number of moles of O₃ is $n_{O_3}=\frac{0.48}{48}=0.01$.

The number of moles of O₂ is $n_{O_2}=0.04$.

The volume of O₂ (volume reduction) is $V_{O_2}=0.04\times 22400=896ml$.