Rate and mechanism of the reaction of Mg and HCl In the reaction between hydrochloric acid and Magnesium the overall reaction is:

$$\ce{Mg + 2HCl -&gt; MgCl2 + H2}$$

My proposed mechanism of reaction is that $\ce{HCl}$ dissociates in $\ce{H2O}$ like so:
$$\ce{H2O + HCl &lt;=&gt; H3O+ + Cl- }$$

And then the $\ce{H3O+}$ oxidises the magnesium like so:

$$\ce{2H3O+ + Mg -&gt; Mg^{2+} + H2O + H2}$$

This, in an experiment to determine the rate of the reaction, measuring the $\ce{H2}$ produced would measure this mechanism, because it is the step in which hydrogen gas is evolved. Presumably these Magnesium ions are then in solution. I would therefore deduce that the step:
$$\ce{2H3O+ + Mg -&gt; Mg^{2+} + H2O + H2}$$
Is the rate-determining step, because it is the step in which magnesium ions are formed and go into solution.
My rate law, deduced from these mechanisms, would look like this:

$$\ce{rate  = k * [H3O+] }$$
Because it is the concentration of the $\ce{H3O+}$ which determines the formation of the Mg ions. From there, I would assume, the $\ce{Mg^{2+}}$ and the $\ce{Cl-}$ would remain in solution, as the reaction is carried out in the acid, obviously consisting of water. 

From this, can I say that the rate law for the reaction

$$\ce{Mg + 2HCl -&gt; MgCl2 + H2}$$

is merely:
$$\ce{rate  = k * [H3O+] }$$
But $\ce{[H3O+]}$ is:
$$\ce{[H3O+] = (K*[H2O][HCl])/[Cl^{-}] }$$
Where K is the equilibrium constant of the acid equilibrium.

So can we write our rate law like:

$$\ce{rate = k*(K[H2O][HCl])/[Cl^{-}] }$$
Where K is the equilibrium constant and k is the rate constant and might look like this (following the Arrhenius equation):
$$\ce{k = A e^{-E_a/(R T)}}$$

To make our rate law like this:
$$\ce{ rate = A e^{-E_a/(R T)} *((K*[H2O][HCl])/[Cl^{-}]) }$$

My question is: Is my rate law correct for this reaction? If not, where have I gone wrong? 

Additionally: This was one of our practicals. We ended up concluding that rate goes up quadratically or exponentially with concentration. This rate law would seem to indicate a linear increase, not exponential. I assume in my rate law there is no contribution from the solid Mg. Obviously in a powder, more surface area is exposed, and a high rate is experienced. Is there a way to factor this into my rate law? 
Thanks for the help! :)
Your rate law should be
$$ r = k_\text{obs} \times [\ce{H3O+}]^2 = A_\text{obs} \exp\left(\frac{E_\text{a, obs}}{RT}\right) \times [\ce{H3O+}]^2\; ,$$
which nicely fits your experimental data of a quadratic dependence on the concentration for the rate.

The exponent is necessary because the reaction is seemingly second-order in the concentration of $\ce{H+}$.