Q:

What can be done to increase the yield of a product in a chemical reaction that has reached equilibrium?

Let's take the following reaction that has reached equilibrium
$$\ce{CaCO3 (s) <=> CaO (s) + CO2 (g)}$$
What can be done to increase the yield of calcium oxide ?
My initial thought was to remove the carbon dioxide produced in the reaction mixture. I understand that this is correct since, according to Le Chatelier's principle, the reaction would shift to increase the yield of the products. 
Why wouldn't adding more calcium carbonate lead to a shift to the products and hence lead to a greater yield of the products? Doesn't more reactant mean more product?

A:

I agree with your initial starting point, that the thermal decomposition of calcium carbonate into calcium oxide and carbon dioxide may be treated as an example of an heterogeneous equilibrium
$$\ce{CaCO3 (s) <=> CaO (s) + CO2 (g)}$$
that still is of relevance today

(source, Hughes et al. in Ind. Eng. Chem. Res. 2004, 43, 5529-5539).
and actually at around $\ce{900 ^\circ{}C}$, the equilibrium pressure of $\ce{CO2}$ over $\ce{CaCO3}$ and $\ce{CaO}$ would equal to $\pu{1 atm}$.
Your first reasoning, while keeping the other parameters of this equilibrium fixed ($T, p$), a removal of $\ce{CO2}$ may increase the yield of $\ce{CaO}$ since this eventually (macroscopically) leads to an exhaustive decomposition of $\ce{CaCO3}$ is correct.

For the second part, you have to balance the general idea about chemical equilibria (which in this case indeed means, keeping other parameters fixed, adding more $\ce{CaCO3}$ as starting material yields more $\ce{CO2}$) with the fact that this particular reaction is heterogeneous.  Looking up the data for molar mass and density, (1, 2, and 3) allows you to estimate the volume one mole of each pure compound under standard conditions occupies:
| compound | molar mass |  density | molar volume |
|          |    [g/mol] | [g/cm^3] |   [cm^3/mol] |
|----------+------------+----------+--------------|
| CaCO3    |     100.09 |    2.711 |        36.92 |
| CaO      |      56.08 |     3.37 |        16.64 |
| CO2      |      44.01 | 0.001977 |     22261.00 |
|----------+------------+----------+--------------|

The value of the molar volume of $\ce{CO2}$ may remind you to the value of the volume occupied by the Ideal Gas at standard conditions ($\pu{22.7 L/mol}$) and demonstrates that the influence of $\ce{CO2}$ on the position of the equilibrium is much more important than the two other of the solids with their -- in comparison -- often negligible molar volumes. Hence, to underline that this equilibrium is this much pressure dependent, it is described by $K_{\text{p}}$, rather than by a normal $K$ value.
The general idea of influencing a chemical equilibrium along "an increase of  partial volume (or partial pressure) of starting material(s) will increase the yield of the product(s)" better full-filled in the case of homogeneous reactions, like the synthesis of ammonia 
$$\ce{N2(g) + 3 H2(g) <=> 2 NH3(g)}. $$

