Calculate the number of moles produced per hour per liter of air? A reaction of importance in the formation of smog is that between ozone and nitrogen monoxide described by

$$
\ce{O3(g) + NO(g)-&gt;O2(g) + NO2(g)}
$$

the rate law for this reaction is:

$$
\text{rate of reaction} = k\left[\ce{O3}\right]\left[\ce{NO}\right]
$$

Given that $k = 2.98 \times 10^6\ \mathrm{M^{–1}\cdot s^{–1}}$ at a certain temperature, calculate the initial reaction rate when $\left[\ce{O3}\right]$ and $\left[\ce{NO}\right]$ remain essentially constant at the values $\left[\ce{O3}\right]_0 = 5.34 \times 10^{–6}\ \mathrm{M}$ and $\left[\ce{NO}\right]_0 = 7.99 \times 10^{–5}\ \mathrm{M}$, owing to continuous production from separate sources.

So, I have found that by using the rate of reaction equation 

$$
\text{the rate of reaction} = $0.00127\ \mathrm{M/s}
$$

Now I’m wondering how to calculate the amount of substance of $\ce{NO2(g)}$ produced per hour per liter of air.

Right now, my idea is that you can take 

$$
\left(1.27\times 10^{-3}\ \mathrm{M}\over\mathrm{s}\right)\left(60\ \mathrm{s}\over{1\ \mathrm{min}}\right)\left(60\ \mathrm{min}\over{1\ \mathrm{h}}\right) = \ce{NO2} \,\text{per hour per liter of air}
$$

Is this the correct way to go about it? I was thinking that the seconds and minutes would cancel leaving just $\mathrm{M/h}$ and then if you multiply by 1 liter it would be $\mathrm{M/(h\times L)}$.
That is the correct way to go about it. In your calculations the rate of the reaction should include L because M is defined as moles/L. (1.27×10^-3 moles/Ls)(3600s/Hour) = 4.572 moles/Lh NO₂ produced.