Question

The conversion of oxygen to ozone occurs to the extent of $15\% $ the mass of ozone that can be prepared from the 67.2 L if oxygen at 1 atm and 273 K will be:
A) 14.4gm
B) 96gm
C) 640gm
D) 64gm

Answer 1

Hint: Remember that 1 mole of any compound will occupy a volume of 22.4 L at Standard Temperature and pressure conditions. The conditions for STP are 273K and 1 atm, only then 1 mole will occupy 22.4 L of volume.

Complete answer: The balanced chemical equation of the conversion of Oxygen to Ozone can be given as:
$3{O_2} \to 2{O_3}$
Which means that 3 moles of dioxygen will produce 2 moles of Ozone.
To find the no. of moles of oxygen given to us we will divide the given volume by 22.4 L
$moles({O_2}) = \dfrac{{Volume}}{{22.4}} = \dfrac{{67.2}}{{22.4}} = 3mol$of dioxygen
Hence, we have been given 3 moles of oxygen which is required for the complete conversion of oxygen to ozone. The condition of 3 moles of dioxygen will produce 2 moles of Ozone is satisfied if the reaction has 100% efficiency. But here we are given that the reaction occurs to about 15% only.
${O_3}$ produced = 2 moles (for 100% efficiency)
${O_3}$ produced = $2 \times \dfrac{{15}}{{100}}mol = 0.3mol$ (for 15% efficiency)
To find the amount of ${O_3}$ in grams, we’ll use the formula: $mass = moles \times {M_{{O_3}}}$
(Consider that the molar mass of ozone is 48g/mol)
The mass of ozone that can be prepared $ = 0.3 \times 48 = 14.4g$
The correct answer is Option (A).

Note:
The alternative approach to the problem can be:
3 moles of dioxygen will produce 2 moles of Ozone
i.e. $3 \times 22.4$L of Oxygen will produce $2 \times 22.4L$ of Ozone. Given that the reaction has 15% efficiency, 67.2L of Oxygen will give: $67.2 \times \dfrac{{15}}{{100}} \times \dfrac{2}{3} = 6.72L$ of ozone.
Mass of 22.4 L of ozone will be 48g then mass of 6.72 L of ozone will be: $\dfrac{{48}}{{22.4}} \times 6.72 = 14.4g$