Application of LeChatelier's principle for reactions with solid products
I wanted to understand application of LeChatelier's principle for reversible reactions of type $\ce{A(g) &lt;=&gt; B(g) + C(s)}$. Because of increase in moles, you can say lower pressure will favor the reaction going right. But does this apply even if one of the products is solid, like in this reaction? If not, why? I need to understand this intuitively, if possible, with less math.
  


Let's get back to the fundamentals of Le Chatelier's principle : 


  When any system at equilibrium is subjected to change in concentration, temperature, volume, or pressure, then the system readjusts itself to counteract (partially) the effect of the applied change and a new equilibrium is established.


Let's assume you have a reaction $\ce{A(g) &lt;=&gt; 2B(g)}$. When lowering the pressure, the system will adjust itself to counteract this, i. e. increase the pressure. This can be achieved by displacing the equilibrium towards the right, as the direct reaction increases the amount of gas molecules.

In your case, however, there is no increase in the amount of gas compounds. Regarding the solid, as it does not participate to the overall pressure in the system (only gases do), it should not be taken into consideration here. Therefore, a change in pressure wouldn't alter the equilibrium state of your reaction.

That's for the math-less part.

You can also understand this concept using the equilibrium constant. For my example, it would be $K_c = \frac{[\ce{B}]^2}{[\ce{A}]}$. Lowering the pressure corresponds to an increase in volume, thus a decrease in both concentrations. This leads to a ratio that is different from the equilibrium constant, which will force the system to a new equilibrium.
$$K_c = \frac{[\ce{B}]^2}{[\ce{A}]} = \frac{\left(\frac{n_\ce{B}}{V}\right)^2}{\frac{n_\ce{A}}{V}} = \frac{n_\ce{B}^2}{n_\ce{A}}\cdot\frac{1}{V}$$
When increasing $V$ to $V'$, 
$$\frac{[\ce{B}]^2}{[\ce{A}]} = \frac{n_\ce{B}^2}{n_\ce{A}}\cdot\frac{1}{V'} &lt; K_c$$
As this ratio is lower than $K_c$, the reaction will evolve towards the formation of products.

For your equation, 
$$K_c = \frac{[\ce{B}]}{[\ce{A}]} = \frac{\frac{n_\ce{B}}{V}}{\frac{n_\ce{A}}{V}} = \frac{n_\ce{B}}{n_\ce{A}}$$
there is no dependance on the volume, thus none either on the pressure.
C is not included as it is a pure solid ($a(\ce{C}) = 1$).


  
  Another related question I have is we talk about reactions having thermodynamic limitations when considering their practical applicability. i.e say a reaction such as above might only yield 20% conversion at industrially practical conditions. But isn't this a bit misleading? As long as you find a way to remove products as soon as they formed, you can push ANY reversible reaction to the right. So for all practical purposes you can say any reversible reaction can be designed to give high conversions (80-100%) if you devise a mechanism to remove products quickly from reaction space. So the thermodynamic limitation is not  really a hard limitation at all.
  


It is true that, if you find a way to remove products from the system, you can reach full conversion of your reagents. But that's actually the catch : it can be very hard to remove those products. Take as an example the Haber-Bosch process : 
$$\ce{N2(g) + 3 H2(g) &lt;=&gt; 2 NH3(g)}$$
As it is a slow reaction, a high temperature is required to get the product in a reasonable time. How would you remove $\ce{NH3}$ from the system, leaving the reagents inside? We then use other ways of increasing the yield (pressure and temperature), but they do not allow a full conversion of the reagents.