Using the mass of calcium carbonate, what volume of carbon dioxide could theoretically be obtained if the reaction progressed to completion? Using the mass of calcium carbonate $m(\ce{CaCO3}) = \pu{0.500 g}$, what volume of carbon dioxide $V(\ce{CO2})$ (in $\pu{mL})$ could theoretically be obtained if the reaction progressed to completion? Assume that the pressure was $\pu{101.3 kPa}$ and the temperature was $\pu{22.5 ^\circ C}$ and that the gas was behaving ideally $(R = \pu{8.314 kPa L mol-1 K-1})$.

I know that I will work with
$$pV = nRT$$
and to bring the $n$ (amount of substance), I have to divide the mass by the molar mass and I assumed that $\ce{CO2}$ mass will be $\pu{0.5 g}$ as well because $\ce{CaCO3}$ will give $\ce{CaO + CO_2}$ so the ratio is $1:1$ so $\ce{CO2}$ mass is $\pu{0.5 g}$, but my problem is that when I bring the amount of substance it should be for $\ce{CO_2}$ or $\ce{CaCO3}$? Because it will differ, $\frac{0.5}{44.01}$ or $\frac{0.5}{100.09}$, that's what I am stuck on.
You are confused about chemical equations. The stoichiometric coefficient (the numbers before a compound) do not indicate the mass of the compound, they indicate the amount of substance. For example
$$\ce{CaCO3 -&gt; CaO + CO2}.$$
This equation means that $\pu{1 mol}$ of calcium carbonate will decompose to give $\pu{1 mol}$ of $\ce{CaO}$ and $\pu{1 mol}$ of $\ce{CO2}$.
It does not mean that if 1 gram calcium carbonate were to decompose, 1 gram of calcium oxide and 1 gram of carbon dioxide will be obtained.
In your question the mass of $\ce{CaCO3}$ is given. From it you can calculate the moles of $\ce{CaCO3}$ involved. Same number of moles of $\ce{CO2}$ will be obtained and you can put that in the ideal gas equation.
Just so you know,
$$n_{\ce{CO_2}} = n_{\ce{CaCO_3}} = \frac{\pu{0.5 g}}{\pu{100.09 g/mol}}.$$
I recommend you study the chapter about the mole concept again.
As a side note: You mentioned in your question that the ratio of products is $1 : 1$. That's actually the ratio of the amounts of substances, not the mass ratio.