Therefore, for every mole of $\mathrm{O}_{2}$ formed, 2 moles of $\mathrm{N}_{2} \mathrm{O}_{5}$ are consumed. So, if the concentration of $\mathrm{O}_{2}$ is $0.1 \mathrm{~mol} \mathrm{~L}^{-1}$, the concentration of $\mathrm{N}_{2} \mathrm{O}_{5}$ consumed is $2
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