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Question

(A) The balanced chemical equation for the conversion of oxygen to ozone is: $3O_{2}(g) \rightarrow 2O_{3}(g)$.Calculate the initial moles of $O_{2}$

Vedclass · Accepted Answer

$14.4$

Vedclass · Answer

(A) The balanced chemical equation for the conversion of oxygen to ozone is: $3O_{2}(g) ightarrow 2O_{3}(g)$.Calculate the initial moles of $O_{2}$ at $STP$ $(1 \ atm, 273 \ K)$: $n(O_{2}) = \frac{67.2 \ L}{22.4 \ L/mol} = 3 \ mol$.Since the conversion occurs to the extent of $15\%$,the moles of $O_{2}$ actually converted are: $3 \ mol 	imes 0.15 = 0.45 \ mol$.According to the stoichiometry,$3 \ mol$ of $O_{2}$ produce $2 \ mol$ of $O_{3}$.Therefore,$0.45 \ mol$ of $O_{2}$ will produce: $n(O_{3}) = \frac{2}{3} 	imes 0.45 \ mol = 0.3 \ mol$.The molar mass of $O_{3}$ is $3 	imes 16 = 48 \ g/mol$.Mass of $O_{3}$ formed $= 0.3 \ mol 	imes 48 \ g/mol = 14.4 \ g$.

The conversion of oxygen to ozone occurs to the extent of $15\%$ only. The mass of ozone that can be prepared from $67.2 \ L$ of oxygen at $1 \ atm$ and $273 \ K$ will be ............ $g$.

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