Heating oxygen in the presence of a catalyst | vedclass

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(C) The balanced chemical equation for the formation of ozone is: $3O_2 \rightarrow 2O_3$.At $STP$,$22.4 \ L$ corresponds to $1 \ mol$ of gas.Number o

Vedclass · Accepted Answer

$480$

Vedclass · Answer

(C) The balanced chemical equation for the formation of ozone is: $3O_2 ightarrow 2O_3$.At $STP$,$22.4 \ L$ corresponds to $1 \ mol$ of gas.Number of moles of $O_3$ produced $= \frac{33.6 \ L}{22.4 \ L/mol} = 1.5 \ mol$.According to the stoichiometry,$2 \ mol$ of $O_3$ are produced from $3 \ mol$ of $O_2$.Therefore,$1.5 \ mol$ of $O_3$ requires: $\frac{3}{2} 	imes 1.5 = 2.25 \ mol$ of $O_2$ (theoretical).Since the conversion efficiency is $15\%$,the actual amount of $O_2$ required is: $\frac{2.25 \ mol}{0.15} = 15 \ mol$ of $O_2$.The molar mass of $O_2$ is $32 \ g/mol$.Mass of $O_2$ required $= 15 \ mol 	imes 32 \ g/mol = 480 \ g$.

Heating oxygen in the presence of a catalyst results in $15\%$ conversion to ozone. How many grams of oxygen are required to produce $33.6 \ L$ of $O_3$ at $STP$?

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