Computing the concentration of a reactant given formation of products At high temperatures nitrogen dioxide decomposes into nitrogen oxide and oxygen gas according to $$\ce{2 NO_2 -&gt; 2 NO + O_2}$$

I have a question now. Let's say at a certain temperature I'm given data for the formation of $\ce{O_2}$. For example:
At $t = 0$ we have $[\ce{O_2}] = 0$ mol/L, at $t = 50$, $[\ce{O_2}] = 7.10 \cdot 10^{-4}$ mol/L, at $t = 100, [\ce{O_2}] = 1.205 \cdot 10^{-3}$ and so on until a certain time.

If I'm given that the initial concentration of $\ce{NO_2}$ is $0.006M$, how can I know what the concentration of $\ce{NO_2}$ is at certain times? I'm asked to determine the order of the reaction on the basis of the given data, and also to present graphically the rate constant.

To do that I need to know if $\ln([\ce{NO_2}])$ versus time $t$ is a straight line. But how to find $[\ce{NO_2}]$ in the first place?
In this reaction, the production of $\ce{O2}$ is tied to the decomposition of $\ce{NO2}$; for each molecule of $\ce{O2}$ that is produced, there have to be 2 molecules of $\ce{NO2}$ that have decomposed. So:

$$2 \Delta n_{\ce{O2}} = -\Delta n_{\ce{NO2}}$$

where $\Delta n$ represents a change in the number of moles. The question does not specify, but I will assume that the volume of the reaction is kept constant; since concentration is amount divided by volume, this means that the same relation holds for concentrations:

$$2\Delta [\ce{O2}] = -\Delta [\ce{NO2}]$$

Since the initial concentration of $\ce{O2}$ is zero, $\Delta [\ce{O2}] = [\ce{O2}]$, and therefore

$$2[\ce{O2}]_t = [\ce{NO2}]_i - [\ce{NO2}]_t $$

$$[\ce{NO2}]_t = [\ce{NO2}]_i - 2[\ce{O2}]_t$$

To use the numbers in the question, at $t = 50$ s we have $[\ce{NO_2}] = (0.006 - 2(7.10\cdot 10^{-4}))$ M = $4.58\cdot 10^{-3}$ M.