If I have the heterogeneous equilibrium

$$\ce{aA(s) + bB(g) <=> cC(g) + dD(g)}$$

I understand that the equilibrium expression is

$$K_c = \frac{[\ce{C}]^c[\ce{D}]^d}{[\ce{B}]^b}$$

and that $\ce{A}$ is left out of the equilibrium expression since it is a pure solid and pure solids/liquids are left out of the equilibrium expression in heterogeneous equilibria.

However, does this mean that adding or removing substance $\ce{A}$ has no effect on the reaction at all? For instance, is it true that adding $\ce{A}$ does not produce more products and removing $\ce{A}$ does not produce more reactants (as Le Chatelier's principle would predict for a reactant included in the equilibrium expression, such as $\ce{B}$)?

On a side note, adding/removing the other substances could potentially affect $[\ce{A}],$ right? For example, adding $\ce{C}$ or $\ce{D}$ would cause $[\ce{A}]$ to increase (because more $\ce{B}$ would be formed and with that more $\ce{A}$ would be formed as well) and adding more of substance $\ce{B}$ would cause $[\ce{A}]$ to decrease (because $\ce{B}$ would be consumed and with it $\ce{A}$ would be consumed as well), correct?

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    $\begingroup$ In order for an equilibrium to exist A must be in excess. $\endgroup$ – MaxW May 26 '19 at 22:53
  • $\begingroup$ @MaxW Why is this? $\endgroup$ – Michael May 27 '19 at 2:13
  • $\begingroup$ "Excess" is not quite the correct word. The idea is more that there needs to be sufficient A so that when the equilibrium is established then there is still some A remaining. // If there is no A remaining then there can't be an equilibrium. In that case A is the limiting reagent. $\endgroup$ – MaxW May 27 '19 at 4:35
  • $\begingroup$ [A] does not change when the amount of solid increases or decreases (the distance of particles in the pure solid is a constant). It will become zero when the solid is used up, otherwise it is a constant. $\endgroup$ – Karsten Theis May 28 '19 at 15:01