Chemical Equilibrium - Le Chatelier's principle If the concentration of reactants is increased such that one is a limiting agent, wouldn't the concentration of products also increase (because excess of a reactant will get carried to the other(product) side  of the equation considering that the limiting agent is in aqueous medium? 

e.g. 
  (Consider 1 mole of each reactant)
   $$\ce{H2O + NH3 &lt;=&gt; NH4+ + OH-}$$

Now, let us take another scenario in which  there are two moles of NH3 and 1 mole of H20. Then concentration of one reactant has increased (from the previous scenario). Wouldn't the concentration of the products also increase because one mole of NH3 does not react and gets carried to the products' side of the equation?

I am confused with how  Le Chatelier's principle is applied in the 2nd scenario. Why does the equilibrium shift to the right if the concentration of the reactants and products increase (if my assumption is right)

If my assumption is wrong (concentration of product increases to balance excess reactant), my concern is such an equation is not balanced. I am unable to reconcile how the Le Chatelier's principle is applied in the 2nd scenario.

(Based on the answers below, I would like to restate my question)
I understand that

$$\ce{2H2O &lt;=&gt; H3O+ + OH-}$$

Now if I add HCl, I have 

$$\ce{H2O + HCl &lt;=&gt; H3O+ + Cl-}$$

The way I interpret this is : the concentration of reactants  is 1 + 1 = 2 moles and the concentration of products is also  1 + 1 = 2 moles. Since the concentration on both sides of the equation is same, why do we say equilibrium shifts left. What am I missing?
For the case you're describing, having left-over product species will likely not cause the reverse reaction to increase.

When you define a reaction equation like: $\ce{aA + bB &lt;=&gt; cC + dD}$, you are explicitly defining A and B as 'reactants' and C and D as 'products'. In the scenario you've described, just because you have excess reactants because you have a limiting concentration of either A or B, doesn't mean the concentration of products will increase. This is because we can think of concentration of products as $[Products](t) = [C](t) + [D](t)$ and not $[Products](t) = [C](t) + [D](t) + [A](t) + [B](t)$.

--

A nice definition of Le Chatlier's principle is the following: "A change in one of the variables that describe a system at equilibrium produces a shift in the position of the equilibrium that counteracts the effect of this change."

If one is to get a little more technical with this, equilibrium is a function of many factors, primarily Temperature (T), Pressure (P), and mixture species (n). This is typically described by thermodynamics by the state property, Gibbs Free Energy, or $\Delta G$.

For the purposes of an example to quantify Le Chatlier's principle, let's look at the differential form of Gibbs Free Energy: $dG = -SdT +VdP + \sum^{n}_{i=1}\mu_idn_i$. If we consider a case where temperature isn't changing (isothermal), and where pressure isn't changing (isobaric), then the Gibbs Free Energy equation is affected by changes in system composition: $dG = \sum^{n}_{i=1}\mu_idn_i$. For an equilibrium case, which means that $\Delta G = 0$, it can be shown that if you expand this expression out for a reaction mechanism (like the one mentioned above), you end up with an expression like: $\frac{\Delta G^o_{T_o}}{RT_o} = -ln(K_{eq})$. This particular equation is defined for standard conditions (0 degC, 1 atm).

--

We're almost to an answer, let's look a little closer at $K_{eq}$. Let's use a gas phase system $\ce{aA + bB &lt;=&gt; cC + dD}$ as an example. If you look up the derivation of the $\frac{\Delta G^o_{T_o}}{RT_o} = -ln(K_{eq})$ in any standard thermodynamics textbook, you will see that $K_{eq}$ is actually equal to the ratio of the mol fractions of each species multiplied by the system pressure: $K_{eq} = \frac{y_C^c y_D^d}{y_A^a y_B^b} P^{c + d - a - b}$. If one assumes that the pressure is fixed and that $a = b = c = d = 1$, then this expression would reduce down to: $K_{eq} = \frac{y_C y_D}{y_A y_B}$.

You can further add detail to this expression by defining the mol fraction in terms of the extent of reaction $\xi$ for a reaction system. 
Where, $n_i = n_{io} + v_in_{io}\xi$--which essentially says that the mols of a given species is equal to inital mols +/- the extent to which that chemical is generated or reacted away (based on the specified chemical of reference $n_{io}$). If you plug this into the expression for $K_{eq}$ you can see that now the equilibrium conversion of a reaction system is dependent on the amount of mols of each species. For the system described above, the total mols $N_{total}$ will cancel out of the expression leaving:

$K_{eq} = \frac{(n_{Ao} + v_An_{io})(n_{Bo} + v_Bn_{io})}{(n_{Co} + v_CN_{io})(n_{Do} + v_Dn_{io})} = exp(-\frac{\Delta G^o_{T_o}}{RT_o})$

For a given temperature and pressure, the Gibbs Free Energy ($\Delta G$) will be a fixed value. Therefore the conditions in the equation (composition of the mixture of reaction system) must balance in such a way that the equilibrium conversion $\xi$ will satisfy the condition. You can see here that as one either varies products (C,D) or reactants (A,B) that the solution for $\xi$ will change. You can generally show that an excess in either products or reactants will shift the system to maximum equilibrium conversion $\xi$. And as such, if one suddenly added or removed products or reactants you would cause a shift in the equilibrium conversion $\xi$.