92 g of NO_{2(g)} and 32 g of O_{2(g)} are | vedclass

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(C) Given: $92 \ g \ NO_2 = \frac{92}{46} = 2 \ mol$ and $32 \ g \ O_2 = \frac{32}{32} = 1 \ mol$.Since $V$ and $T$ are constant,$n \propto P$,so $n_f

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$25$

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(C) Given: $92 \ g \ NO_2 = \frac{92}{46} = 2 \ mol$ and $32 \ g \ O_2 = \frac{32}{32} = 1 \ mol$.Since $V$ and $T$ are constant,$n \propto P$,so $n_f = \frac{7}{8} n_i$.Initial moles $n_i = 2 + 1 = 3 \ mol$.Reaction: $4NO_{2(g)} + O_{2(g)} 	o 2N_2O_{5(g)}$.At equilibrium: $NO_2 = 2-4x$,$O_2 = 1-x$,$N_2O_5 = 2x$.Final moles $n_f = (2-4x) + (1-x) + 2x = 3-3x$.Given $n_f = \frac{7}{8} 	imes 3 = 2.625$.$3 - 3x = 2.625$ $\Rightarrow 3x = 0.375$ $\Rightarrow x = 0.125 = \frac{1}{8}$.Actual yield of $N_2O_5 = 2x = 2 	imes \frac{1}{8} = 0.25 \ mol$.Theoretical yield: $NO_2$ is the limiting reagent ($2 \ mol \ NO_2$ produces $1 \ mol \ N_2O_5$).Percentage yield $= \frac{0.25}{1} 	imes 100 = 25 \%$.

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